Category: Abstract Algebra

Generalised Fractions
Let $R$ be a commutative ring. Fix a nonempty subset

$D \subseteq R^* =: R\backslash \{0\}$

that is closed under multiplication.
- For any $u \in D$ and $v \in R^*$ , $uv \neq 0$ and $vu \neq 0$ .
- $D$ is closed under multiplication.
For instance $D = \mathrm{U}(R) \subseteq R^*$ works. For a concrete example,

$D = \{-1,1\} = \mathrm{U}(\mathbb Z) \subseteq \mathbb Z^*$

works.

Problem 1. Show that the relation $\sim$ on $R \times D$ defined by

$(r_1,d_1) \sim (r_2,d_2) \quad \iff \quad r_1 d_2 = r_2 d_1.$

In this case, we denote $r/d \equiv [(r, d)]$ .

(Click for Solution)

Solution. Reflexivity and symmetry are obvious. For transitivity, suppose

$\begin{aligned} (r_1,d_1) \sim (r_2,d_2) \quad &\iff \quad r_1 d_2 = r_2 d_1, \\ (r_2,d_2) \sim (r_3,d_3) \quad &\iff \quad r_2 d_3 = r_3 d_2. \end{aligned}$

Then

$\begin{aligned} r_1 d_3 d_2 &= r_1 d_2 d_3 \\ &= r_2 d_1 d_3 \\ &= r_2 d_3 d_1 \\ &= r_3 d_2 d_1 = r_3 d_1 d_2, \end{aligned}$

so that $(r_1 d_3 - r_3 d_1) d_2 = 0$ . Since $d_2 \in D$ , we must have

$r_1 d_3 - r_3 d_1 = 0 \quad \Rightarrow \quad (r_1,d_1) \sim (r_3,d_3).$

Problem 2. Show that $(R \times D)/{\sim}$ can be equipped with a ring structure that turns $(R \times D)/{\sim}$ into a commutative ring with a multiplicative identity, and contains $R$ as a sub-ring. Furthermore, show that if $R$ is an integral domain, $Q(R) := (R \times R^*)/{\sim}$ forms a field.

(Click for Solution)

Solution. Intuitively, we would like elements of the form $r/d$ to match the addition and multiplication of rational numbers. Hence, define

$\begin{aligned} (r_1/d_1) + (r_2/d_2) &:= (r_1 d_2 + r_2 d_1)/(d_1 d_2), \\ (r_1/ d_1) \cdot (r_2 / d_2) &:= (r_1 r_2)/(d_1 d_2), \end{aligned}$

both of which can be checked to be well-defined since $d_1 d_2 \neq 0$ . It is not hard to check that the ring axioms hold with additive identity $0/d$ and multiplicative identity $d/d =: 1$ . To justify this claim, we observe that for any $r/d \in (R \times D)/{\sim}$ ,

$r/d \cdot d/d = (rd)/(d \cdot d).$

Since $r \cdot (d \cdot d) = (r \cdot d) \cdot d$ , we have $(r,d) \sim (rd, d\cdot d)$ , so that

$r/d \cdot d/d = (rd)/(d \cdot d) = r/d,$

as required.

Furthermore, define the map

$f : R \to (R \times D)/{\sim},\quad f(r) := rd/d$

for any $d \in D$ . We remark that the choice of $d$ does not affect the result since for any $d_1, d_2 \in D$ ,

$\begin{aligned} (rd_1)d_2 = (rd_2)d_1 \quad &\Rightarrow \quad (rd_1, d_1) \sim (rd_2, d_2) \\ &\Rightarrow \quad [(rd_1, d_1)] = [(rd_2, d_2)] \\ &\Rightarrow \quad rd_1/d_1 = rd_2/d_2. \end{aligned}$

Hence, it remains to check that $f$ is an injective ring homomorphism, which is not difficult, since

$\begin{aligned} f(r+s) &= (r+s)d/d \\ &= (rd+sd)/d \\ &= ((rd+sd)d)/(d\cdot d) \\ &= (rd \cdot d+sd \cdot d)/(d\cdot d) \\ &= (rd)/d + (sd)/d \\ &= f(r) + f(s) \end{aligned}$

and

$\begin{aligned} f(r \cdot s) &= (r \cdot s)d/d \\ &= ((r \cdot s)d \cdot d)/(d \cdot d) \\ &= ((rd) \cdot (sd))/(d \cdot d) \\ &= (rd)/d \cdot (sd)/d \\ &= f(r) \cdot f(s) \end{aligned}$

and

$f(r) = 0 \quad \Rightarrow \quad rd/d = 0 \quad \Rightarrow \quad r = 0.$

To justify the last implication, use $0 = 0/d$ so that

$\begin{aligned} rd/d = 0/d \quad &\Rightarrow \quad (rd, d) \sim (0, d) \\ &\Rightarrow \quad (rd) \cdot d = 0 \cdot d \\ &\Rightarrow \quad r \cdot (d \cdot d) = 0. \end{aligned}$

Since $d\cdot d \neq 0$ , we must have $r = 0$ , as required.

Finally, suppose $R$ is an integral domain. Fix any nonzero $r/d \in (R \times R^*)/{\sim}$ . If $r = 0$ , then $r/d = 0/d = 0$ , a contradiction. Therefore, $r \neq 0$ so that $r \in R^*$ . Then $d/r = (r/d)^{-1} \in R \times R^*$ , since

$r/d \cdot d/r = (rd)/(dr) = (rd)/(rd) = 1,$

as required.

Definition 1. For any integral domain $R$ , we call $Q(R) := (R \times R^*)/{\sim}$ the fraction field of $R$ .

Example 1. $\mathbb Q = Q(\mathbb Z)$ forms a field.

Recall that we have defined $i :=\sqrt{-1}$ elsewhere.

For any ring, define $R(i) := R + Ri$ .

Example 2. $\mathbb Q(i) = Q(\mathbb Z(i))$ forms a field.

—Joel Kindiak, 1 Feb 26, 2103H
May 7, 2026
Revisiting Factorisation
Much of our discussion in abstract algebra amounts to jacking our understanding of the integers to the maximum limit, and articulating the “movement” within the integers that match the “movement” in possibly other sets (or in this part of the discussion, integral domains).

One property that strikes us must be the unique factorisation feature that the integers enjoy. We call integral domains with $1$ that share this property unique factorisation domains (UFDs).

Henceforth, let $R$ denote an integral domain with $1$ .

Lemma 1. Call $u \in R$ a unit if it has a multiplicative inverse in $R$ . Write $a \sim b$ to mean that there exists a unit $u$ (i.e. an element in $R$ with a multiplicative inverse) such that $a = ub$ . Then

$\mathrm U(R) := \{u \in R : u\ \text{is a unit}\}$

forms a multiplicative group, and $\sim$ forms an equivalence relation on $R$ .

Proof. Straightforward exercise.

Recall that an integral domain is a principal ideal domain (PID) if all of its ideals take the form $\langle r \rangle$ . We claim that PIDs are UFDs. But in order to justify this ambitious claim, we need to develop some nomenclature surrounding UFDs.

Definition 1. We call $R$ a unique factorisation domain (UFD) if for any non-unit $x \in R$ , there exist irreducible elements $p_1,\dots, p_n \in R$ such that

$\displaystyle x = \prod_{i=1}^n p_i.$

Furthermore, for any irreducible elements $q_1,\dots, q_m \in R$ such that

$\displaystyle x = \prod_{j=1}^m q_j,$

we have $m = n$ and after re-labelling, $p_i \sim q_i$ .

Recall that a non-unit $p$ is irreducible if whenever $p = xy$ , at least (and therefore, exactly) one of $x$ or $y$ is a unit.

We’re first going to check that UFDs do carry integer-esque properties with regards to factorisation. Write $a \mid b$ to mean that there exists $q \in R$ such that $b = qa$ .

Lemma 2. Call a nonzero element $p \in R$ prime if $\langle p \rangle \subseteq R$ is a prime ideal. If $R$ is a UFD, then any element $x \neq 0$ is prime if and only if it is irreducible.

Proof. $(\Rightarrow)$ Fix $x \neq 0$ and suppose $x$ is prime. Write $x = uv$ for irreducible elements $u,v$ without loss of generality. Since $x \mid uv$ and $x$ is prime, $x \mid u$ without loss of generality. Write $u = wx$ for some $w \in R$ so that $x = (wx)v = (wv)x$ . Since $R$ is an integral domain, $wv = 1$ . Hence, $v$ is a unit. Similarly, if $x \mid v$ , then $u$ is a unit. Therefore, $x$ is irreducible.

$(\Leftarrow)$ . Suppose $x$ is irreducible. Fix $uv \in \langle p \rangle$ and write $uv = wp$ for some $w \in R$ . Find unique irreducible elements $p_1,\dots p_{m+m}$ such that

$\displaystyle u = \prod_{i=1}^m p_i,\quad v = \prod_{j=1}^n p_{m+j}$

so that

$\displaystyle \prod_{i=1}^{m+n} p_i = u \cdot v = w \cdot p.$

Since $p$ is irreducible, we must have $p_1 \sim p$ without loss of generality. Write $p_i = pz$ for some unit $z$ , so that

$\displaystyle u = \prod_{i=1}^m p_i = p_1 \cdot \prod_{i=2}^m p_i = p \cdot z \cdot \prod_{i=2}^m p_i \in \langle p \rangle.$

Lemma 3. Suppose $R$ is a UFD. Given $a, b \in R$ , there exists some $d \in R$ satisfying the greatest common divisor properties:
- $d \mid a$ and $d \mid b$
- If $c$ satisfies the property above, then $c \mid d$ .
- If $d'$ satisfies the two properties above, then $d \sim d'$ .
Proof. Find irreducible elements $p_1,\dots, p_n, q_1,\dots, q_n$ such that

$\displaystyle a = \prod_{i=1}^n p_i^{\alpha_i}, \quad b = \prod_{i=1}^n q_i^{\beta_i}.$

where $\alpha_i \geq 0, \beta_i \geq 0$ , and $p_i^0 := 1, q_i^0:= 1$ . Suppose that $p_i \sim q_i$ for each $i$ . Define

$\displaystyle d := \prod_{i=1}^n p_i^{\min\{ \alpha_i, \beta_i \}}.$

It is obvious that $d \mid a$ . For $d \mid b$ , find units $u_i$ such that $q_i = p_i u_i$ . Note that $u_i^{\beta_i}$ will always be a limit, so that

$\displaystyle d \mid \prod_{i=1}^n u_i^{\beta_i} \cdot \prod_{i=1}^n p_i^{\beta_i} = \prod_{i=1}^n q_i^{\beta_i} = b.$

Now suppose $c \mid a$ and $c \mid b$ . Write

$\displaystyle c = \prod_{i=1}^n p_i^{\gamma_i},\quad \gamma_i \geq 0.$

Then $\gamma_i \leq \min\{\alpha_i,\beta_i\}$ , so that $c \mid d$ .

Finally, given $d'$ satisfying the first two properties, we must have $d \mid d'$ and $d' \mid d$ . Find elements $u,v \in R$ such that $d = ud'$ and $d' = vd$ . Substituting, $d = (uv)d$ . Since $R$ is an integral domain with $d \neq 0$ , we must have $uv = 1$ , so that $u,v$ are units, and hence, $d \sim d'$ .

Lemma 4. Suppose $R$ is a PID. Fix a collection $\{I_n\}$ of non-decreasing ideals, i.e. $I_n \subseteq I_{n+1}$ for any $n \in \mathbb N$ . Then there exists some $N \in \mathbb N$ such that for $k > N$ , $I_k = I_N$ . In this case, we say that $\{I_n\}$ satisfies the ascending chain condition.

Proof. Write $I_n = \langle x_n \rangle$ for each $n$ . Define $I := \bigcup_{n \in \mathbb N} I_n$ . It is not hard to check that $I \subseteq R$ is an ideal. Since $R$ is a PID, there exists $x \in R$ such that $I = \langle x \rangle$ . Since $x \in I = \bigcup_{n \in \mathbb N} I_n$ , we have $x \in I_N$ for some $N \in \mathbb N$ , so that $I = \langle x \rangle = I_N$ . Then for any $k > N$ , $I_N \subseteq I_k \subseteq I = I_N$ .

Theorem 1. If $R$ is a PID, then $R$ is a UFD.

Proof. Suppose for a contradiction instead that $R$ is not a UFD. Then there exists $x_0 \in R$ that cannot be factorised into irreducible elements. In particular, $x_0$ is not irreducible. Therefore, we can find non-units $x_1, y_1 \in R$ such that $x_0 = x_1 y_1 \in \langle x_1\rangle$ .

Inductively, since $x_k$ is not irreducible, there exist non-units $x_{k+1}, y_{k+1} \in R$ such that $x_k = x_{k+1} y_{k+1} \in \langle x_{k+1}\rangle$ . Since $\langle x_n \rangle \subseteq \langle x_{n+1} \rangle$ for each $n$ , we have constructed a collection $\{ \langle x_n \rangle \}$ of non-decreasing ideals, by Lemma 4, there exists $N \in \mathbb N$ such that $\langle x_k \rangle = \langle x_N \rangle$ for $k > N$ .

In particular, $x_{N+1} \sim x_N$ , which means there exists a unit $u \in R$ such that $x_{N+1} = ux_N$ . On the other hand, by construction, $x_N = x_{N+1} y_{N+1}$ , where $x_{N+1}, y_{N+1}$ are not units. Hence,

$x_{N+1} = (u y_{N+1}) \cdot x_{N+1}.$

Since $R$ is an integral domain, we must have $u y_{N+1} = 1$ , implying that $y_{N+1}$ is a unit, a contradiction.

Therefore, $R$ must be a UFD.

Corollary 1. Let $R$ be an integral domain. If $R$ is a Euclidean domain, then it is a PID, and hence, a UFD.

Why bother with factorisation? Because exceedingly early in our mathematical journey, before even entering university, we factorised polynomials again and again and again. In particular, the real-valued turn out to be a Euclidean domain, and hence becomes a PID and a UFD. These constructions become exceedingly useful when discussing field extensions as preludes to Galois theory.

And so, we turn our attention to the polynomials once again, from the ring-theoretic perspective.

—Joel Kindiak, 30 Jan 26, 1112H
May 5, 2026
Commutative Ring Zoo
Recall that $\mathbb Z$ is a ring that is commutative, and contains a multiplicative identity $1$ . Commutative rings have many rich sub-structures, and the integers actually help us uncover some of these results.

Firstly, we note that given integers $m,n$ , if $m \neq 0$ and $n \neq 0$ , then $mn \neq 0$ . This property doesn’t always hold; in the commutative ring $\mathbb Z_6$ , where elements take of the form $[x] \equiv x + 6\mathbb Z$ and whose identity element is $[0]$ , we have $[2] \neq 0$ and $[3] \neq 0$ but

$[2] \cdot [3] = [2 \cdot 3] = [6] = [0].$

We will revisit these rings later on. But for now, we can, and should, appreciate the specialness of multiplying nonzero numbers.

Henceforth, let $R$ be a commutative ring.

Definition 1. We say that $R$ is an integral domain (taking reference from the integers), if $R^{*} := R \backslash \{0\}$ is closed under multiplication. Equivalently, if $x,y \in R$ , then

$xy \neq 0 \quad \Rightarrow \quad x \neq 0\quad \lor \quad y \neq 0.$

Theorem 1. Let $I \subseteq R$ be an ideal. Then $R/I$ forms an integral domain if and only if $R\backslash I$ is closed under multiplication, i.e. for any $x,y \in R$ ,

$x \notin I\quad \wedge \quad y \notin I \quad \Rightarrow \quad x \cdot y \notin I.$

Proof. $(\Rightarrow)$ Suppose $R/I$ forms an integral domain. Fix $x, y \in R$ and suppose $x\notin I$ and $y \notin I$ . Since $R/I$ forms an integral domain,

$[x] \neq [0],\quad [y] \neq [0] \quad \Rightarrow \quad [x \cdot y] = [x] \cdot [y] \neq [0] = I.$

Therefore, $x \cdot y \notin I$ , as required.

$(\Leftarrow)$ Suppose instead that $R\backslash I$ is closed under multiplication. Fix nonzero $[x], [y] \in R/I$ . Then $x \in R \backslash I$ and $y \in R \backslash I$ , so that

$x \cdot y \in R \backslash I \quad \Rightarrow \quad [x] \cdot [y] = [x \cdot y] \neq [0].$

Therefore, $R/I$ forms an integral domain.

Definition 2. We call an ideal $I \subseteq R$ prime if $R \backslash I$ is closed under multiplication.

Equivalently, if $x,y \in R$ , then

$xy \notin I \quad \Rightarrow \quad x \notin I\quad \lor \quad y \notin I.$

Equivalently again, if $x,y \in R$ , then

$xy \in I \quad \Rightarrow \quad x \in I\quad \lor \quad y \in I.$

By Theorem 1, $R/I$ forms an integral domain if and only if $I \subseteq R$ is a prime ideal.

Example 1. By Definition 2, $R \cong R/\{0\}$ is an integral domain if and only if $\{0\} \subseteq R$ is a prime ideal.

Example 2. For any number $n \in \mathbb N$ , $n \mathbb Z \subseteq \mathbb Z$ is a prime ideal if and only if $n$ is prime. By Theorem 1, $\mathbb Z_p = \mathbb Z/p\mathbb Z$ forms an integral domain if and only if $p$ is prime.

Proof. Fix $x, y \in \mathbb Z$ .

$(\Rightarrow)$ Suppose $n$ is prime. By Euclid’s lemma,

$\begin{aligned} xy \in n \mathbb Z \quad & \Rightarrow \quad n \mid xy \\ &\Rightarrow \quad n \mid x \quad \lor \quad n \mid y \\ &\Rightarrow \quad x \in n \mathbb Z\quad \lor \quad y \in n\mathbb Z, \end{aligned}$

so that $n \mathbb Z \subseteq \mathbb Z$ is prime.

$(\Leftarrow)$ Suppose $n$ is not prime. Then there exists two positive integers $1 < r,s < n$ such that $r \cdot s \in n \mathbb Z$ . By construction,

$\begin{aligned} r \nmid n \mathbb Z \quad &\Rightarrow \quad r \in \mathbb Z \backslash n\mathbb Z, \\ s \nmid n \mathbb Z \quad &\Rightarrow \quad s \in \mathbb Z \backslash n\mathbb Z. \end{aligned}$

Hence, $r,s \in \mathbb Z \backslash n\mathbb Z$ but $r \cdot s \in n \mathbb Z$ . Hence, $\mathbb Z \backslash n\mathbb Z$ is not closed under multiplication.

Lemma 1. Let $p$ be prime. Suppose $p \mathbb Z \subseteq I \subseteq \mathbb Z$ as ideals. Then $I = p\mathbb Z$ or $I = \mathbb Z$ .

Proof. Suppose $I \neq p\mathbb Z$ . Fix $q \in I \backslash p\mathbb Z$ . Then $p \nmid q$ implies that $\gcd(p, q) = 1$ , so that

$p\mathbb Z + q \mathbb Z = \mathbb Z.$

If $q \in I$ , then $q \mathbb Z \subseteq I$ . In particular,

$\mathbb Z \subseteq p\mathbb Z + q \mathbb Z \subseteq I + I \subseteq I \subseteq \mathbb Z.$

Therefore, $I = \mathbb Z$ .

Definition 3. We call an ideal $I \subsetneq R$ maximal if given the ideals $I \subseteq J \subseteq R$ , $J = I$ or $J = R$ .

Suppose furthermore that $R$ has a multiplicative identity $1$ .

Theorem 2. Let $I \subseteq R$ be an ideal. Then $I$ is maximal if and only if every nonzero element in $R/I$ has a multiplicative inverse. In this case, we call $R/I$ a field.

Proof. $(\Rightarrow)$ Fix a nonzero element $[x] \in R/I$ . Since $[x] \neq [0] = I$ , we have $x \notin I$ . Define the set

$J := \{z + xy : z \in I, y \in R\} \supsetneq I.$

It is not hard to check that $J \subseteq R$ as an ideal. Since $I$ is maximal, $J = R$ . In particular, there exist $z \in J$ and $y \in R$ such that

$u+ xy = 1 \quad \Rightarrow \quad xy \in 1 + I.$

Since $R$ is commutative, so is $R/I$ , and

$[x] \cdot [y] = [y] \cdot [x] = [1].$

$(\Leftarrow)$ Fix any ideal $J \subseteq R$ such that $I \subsetneq J$ . Fix $x \in J \backslash I$ . Since $R/I$ is a field, there exists $y \in R\backslash I$ such that

$[xy] = [x] \cdot [y] = [1].$

Hence, $xy - 1 \in J$ implies that

$1 = xy - (xy - 1) \in J.$

Hence,

$R = \langle 1 \rangle \subseteq J \subseteq R,$

implying that $I$ is maximal, as required.

Corollary 1. A maximal ideal $I \subseteq R$ is always prime.

Proof. Fix an ideal $I \subseteq R$ . Suppose $I$ is maximal. By Theorem 2, $R/I$ is a field. By definition of a field, $R/I$ is an integral domain. By Theorem 1 and Definition 2, $I$ is prime.

By Lemma 1, proper ideals of $\mathbb Z$ are maximal if and only if they are prime. This observation hints the specialness of $\mathbb Z$ that deems it a ring par excellence. One crucial observation is that $\mathbb Z$ contains a division or a Euclidean algorithm: given integers $a, b$ and $a \neq 0$ , there exist unique integers $q, r$ such that

$b = aq + r\quad \text{and} \quad (r = 0\quad \text{or}\quad |r| < |a|).$

We generalise this idea into that of a Euclidean domain.

Definition 4. We call an integral domain $R$ with multiplicative identity $1$ a Euclidean domain if there exists a non-negative function (called a norm) $f : R \to \mathbb N_0$ such that for any $a, b \in R$ and $a \neq 0$ , there exist unique $q, r \in R$ such that

$b = aq + r\quad \text{and} \quad (r = 0\quad \text{or}\quad f(r) < f(a)).$

In the case $R = \mathbb Z$ and $f( \cdot ) = | \cdot |: \mathbb Z \to \mathbb N_0$ , we recover the usual Euclidean algorithm.

A crucial example of a non-integers Euclidean domain would be the ring of polynomials with real coefficients. We will explore this ring in greater detail in the future.

Theorem 3. Let $R$ be a Euclidean domain. For ideal $I \subseteq R$ , there exists $x \in R$ such that $I = \langle x \rangle$ . In this case, we call $R$ a principal ideal domain, abbreviated PID.

Proof. Denote the norm in $R$ by $f$ . Fix an ideal $I \subseteq R$ . If $I = \{0\}$ , then $I = \langle 0 \rangle$ . If $I = R$ , then $I = \langle 1 \rangle$ . Suppose $\{0\} \subsetneq I \subsetneq R$ (i.e. $I$ is nontrivial).

By the well-ordering principle, the non-empty set $f(I) \subseteq \mathbb N_0$ has a minimum element $f(x)$ for some $x \in I$ , so that $\langle x \rangle \subseteq I$ . We claim that $(\supseteq)$ holds. Fix $y \in I$ . Since $R$ is a Euclidean domain, there exist unique $q,r \in R$ such that

$y = qx + r\quad \text{and} \quad (r = 0\quad \text{or}\quad f(r) < f(x)).$

Since $f(x)$ is minimal, we must have $f(r) \geq f(x)$ , so that $r = 0$ . Therefore, $y = qx \in \langle x \rangle$ , establishing $I \subseteq \langle x \rangle$ .

Definition 5. A ring $R$ possesses the cancellation law if for any $x,y,z \in R$ with $z \neq 0$ ,

$xz = yz \quad \Rightarrow \quad x=y.$

Lemma 2. A commutative ring is an integral domain if and only if it possesses the cancellation law.

Proof. $(\Rightarrow)$ If $xz = yz$ , then $(x-y) z = 0$ . Hence, $x-y = 0$ or $z = 0$ . Since the latter is impossible, we must have $x-y = 0 \Rightarrow x= y$ .

$(\Leftarrow)$ Suppose $xy = yx = 0$ . If $x = 0$ , we are done. Otherwise, $x \neq 0$ , and by the cancellation law, $y = 0$ .

Theorem 4. Let $R$ be a PID. For any nontrivial ideal $I \subseteq R$ , if $I$ is prime, then $I$ is maximal.

Proof. Fix any nontrivial prime ideal $I = \langle x \rangle$ , $x \neq 0$ . Fix any ideal $J$ such that $\langle x \rangle \subsetneq J \subseteq R$ . Since $R$ is a PID, there exists $y \in R\backslash \langle x \rangle$ such that $J = \langle y\rangle$ . We claim that $R = \langle y \rangle$ .

Since $R = \langle 1 \rangle$ , it suffices to check that $1 \in \langle y \rangle$ . Since $x \in \langle y \rangle$ , there exists $u \in R$ such that $x = uy$ . Since $uy = x \in \langle x \rangle$ and $\langle x \rangle$ is prime, either $u \in \langle x \rangle$ or $y \in \langle x \rangle$ . By hypothesis, the latter cannot hold, so we must have $u \in \langle x \rangle$ . Hence, there exists $v \in R$ such that $u = vx$ , so that $x =(vx)y = x(vy)$ . By the cancellation law, $1 = vy \in \langle y\rangle$ .

Definition 6. An element $x \in R^*$ is a unit if it has a multiplicative inverse. An element $x \in R^*$ is irreducible if $x$ is not a unit, and if $x = uv$ for some $u, v \in R$ , at least one of $u, v$ is a unit.

Theorem 5. Let $R$ be a PID and fix $x \in R^*$ . The following are equivalent:
- $\langle x \rangle$ is maximal.
- $\langle x \rangle$ is prime.
- $x$ is irreducible.
Proof. Corollary 1 and Theorem 4 establish the equivalence of the first two points. For the third point, we first suppose $\langle x \rangle$ is prime. Suppose $x = uv$ for some $u, v \in R$ . Since $uv \in \langle x \rangle$ , we must have $u \in \langle x \rangle$ or $v \in \langle x \rangle$ . Suppose the latter without loss of generality, so that $v = wx$ for some $w \in R$ . Then

$x=uv \quad \Rightarrow \quad x = u(wx) = (uw)x.$

Since $R$ is an integral domain, by the cancellation law, $uw = 1$ . Therefore, $u$ is a unit. Similarly, if $u \in \langle x \rangle$ , then $x = vu$ implies that $v$ is a unit. In either case, we have proven that $x$ is irreducible.

Now suppose instead that $x$ is irreducible. We claim that $\langle x \rangle$ is maximal. Fix $y \in R\backslash \langle x \rangle$ such that $\langle x \rangle \subsetneq \langle y \rangle \subseteq R$ . Since $x \in \langle y \rangle$ , find $u \in R$ such that $x = uy$ . Since $x$ is irreducible, either $u$ is a unit or $y$ is a unit. If $u$ is a unit, denote its inverse by $u^{-1}$ , so that $y = u^{-1}x \in \langle x \rangle$ , a contradiction. Hence, $y$ is a unit. Denote its inverse by $y^{-1}$ , so that $1 = y^{-1}y \in \langle y \rangle$ . Hence, $R \subseteq \langle y \rangle \subseteq R$ . Since $\langle y \rangle = R$ , we conclude that $\langle x \rangle$ is maximal.

We conclude our commutative ring zoo tour by defining the unique factorisation domain, which extends our ideas of unique factorisation in $\mathbb Z$ to plausibly other kinds of rings.

Definition 7. Given $a, b \in R$ , write $a \sim b$ if there exists a unit $u \in R$ such that $a = ub$ .

We leave it as an exercise to check that $\sim$ forms an equivalence relation on $R$ . Denote elements of $R/{\sim}$ by $\bar{a} := \{b \in R : a \sim b\}$ .

Definition 8. An integral domain $R$ is a unique factorisation domain (UFD) if for any $x \in R$ , there exist unique elements $\bar{p}_1, \dots, \bar{p}_n \in R/{\sim}$ such that

$\displaystyle \bar x = \prod_{i=1}^n \bar{p}_i.$

Theorem 6. If $R$ is a PID, then $R$ is a UFD.

Proof. Stay tuned for the next post.

—Joel Kindiak, 28 Jan 26, 2310H
May 4, 2026
Chinese Mathematics
In this post, we use the notion of ideals to prove the Chinese remainder theorem.

Theorem 1 (Chinese Remainder Theorem). Let $a,b,m,n$ be integers. If $\gcd(m, n) = 1$ , then there exists a unique integer $0 \leq x < mn$ such that

$x \equiv_m a,\quad x \equiv_n b.$

Problem 1. Given rings $R, S$ , define the operations $+, \cdot$ pointwise:

$\begin{aligned} (r_1, s_1) + (r_2, s_2) &:= (r_1 + r_2, s_1 + s_2), \\ (r_1, s_1) \cdot (r_2, s_2) &:= (r_1 \cdot r_2, s_1 \cdot s_2). \end{aligned}$

Show that

$R \oplus S := \{(r,s) : r \in R, s \in S\}$

forms a ring. More generally, inductively define

$\displaystyle \bigoplus_{i=1}^n R_i := \bigoplus_{i=1}^{n-1} R_i \oplus R_n.$

Solution. Straightforward verification of ring axioms.

Let $R$ be a ring and $I, J \subseteq R$ be ideals.

Problem 2. Show that the map $\phi : R \to R/I \oplus R/J$ defined by

$\phi(x) := (x + I, x+ J)$

is a well-defined ring homomorphism. Evaluate $\ker(\phi)$ .

(Click for Solution)

Solution. Since $I, J$ are ideals, the quotient sets $R/I, R/J$ have natural ring structures. By Problem 1, the product $R/I \times R/J$ has a natural ring structure, and

$\begin{aligned} \phi(x \cdot y ) &= ((x \cdot y) + I, (x \cdot y) + J) \\ &= ((x + I) \cdot (y + I), (x+J) \cdot (y+J)) \\ &= (x+ I, x_J) \cdot (y + I, y + J) \\ &= \phi(x) \cdot \phi(y). \end{aligned}$

Replace $\cdot$ with $+$ to conclude that $\phi$ is a ring homomorphism.

To evaluate $\ker(\phi)$ , fix $u \in \ker(\phi)$ so that

$\phi(u) = (u + I, u + J) = (I, J).$

Then $u \in I$ and $u \in J$ , so that $u \in I \cap J$ . It is clear that $I \cap J \subseteq \ker(\phi)$ , so that $\ker(\phi) = I \cap J$ .

Problem 3. Fix $K \subseteq R$ . Show that there exists a unique ideal $\langle K \rangle \subseteq R$ such that
- $\langle K \rangle$ contains $S$ , and
- for any ideal $L \subseteq R$ that contains $S$ , $L$ contains $\langle K \rangle$ .
(Click for Solution)

Solution. Define

$\mathcal K:= \{L : L \subseteq R\ \text{ideal},\ S \subseteq L\}.$

It is clear that $R \in \mathcal K$ , so that $\mathcal K \neq \emptyset$ . Define $\langle K \rangle := \bigcap_{L \in \mathcal K} L$ . We claim that $(K)$ satisfies the desired properties.

To check that $\langle K \rangle$ is an ideal, we note that for any $x \in R$ and $z \in K$ and ideal $L \in \mathcal K$ , $xz \in L$ . Therefore, $xz \in \bigcap_{L \in \mathcal K}L = \langle K \rangle$ . The right-sided ideal property holds similarly. For the second claim, it is obvious that $\langle K \rangle = \bigcap_{L_0 \in \mathcal K} L_0 \subseteq L$ .

Define $I \cdot J := \langle \{xy : x \in I, y \in J\} \rangle$ .

Problem 4. Show that

$\displaystyle I \cdot J = \left\{ \sum_{k=1}^n x_k y_k : n \in \mathbb N, x_k \in I, y_k \in J \right\}.$

Deduce that $I \cdot J \subseteq I \cap J$ .

(Click for Solution)

Solution. For the first claim, the direction $(\supseteq)$ is trivial.

For the direction $(\subseteq)$ , since the set

$\displaystyle K := \left\{ \sum_{k=1}^n x_k y_k : n \in \mathbb N, x_k \in I, y_k \in J \right\} \subseteq R$

contains the set $\{xy : x \in I, y \in J\}$ , by Problem 3, it suffices to check that it is an ideal. Fix $w \in R$ and $z \in K$ . Write

$\displaystyle z = \sum_{k=1}^n x_k y_k.$

Since $I$ is an ideal, $wx_k \in I$ for each $k$ , so that

$\displaystyle wz = w \left( \sum_{k=1}^n x_k y_k \right) = \sum_{k=1}^n wx_k \cdot y_k \in K.$

The right-sided ideal property holds similarly. Therefore, $K$ is an ideal, as required. For the second claim, each $x_k y_k \in I \cap J$ , and since the latter is an ideal, the result follows.

Problem 5. Given that $I + J = R$ , show that the map $\phi$ defined in Problem 2 is surjective. If furthermore that:
- $R$ is commutative, and
- $1 \in R$ ,
show that $R/(I \cdot J) \cong R/ I \oplus R/ J$ .

(Click for Solution)

Solution. Fix $(x + I, y + J) \in R/ I \oplus R/ J$ .

We need to cook up $z \in R$ such that $z - x \in I$ and $z - y \in J$ , so that

$\phi(z) = (z + I , z + J) = (x + I, y + J).$

Since $I + J = R$ and $x - y \in R$ , there exists $u \in I, v \in J$ such that

$x - y = u + v \quad \Rightarrow \quad x + (-u) = y + v.$

Define

$z := x + (-u) = y+v.$

Then $z-x = -u \in I$ and $z - y = v \in J$ , as required.

By Problem 4, $I \cdot J \subseteq I \cap J$ . If $1 \in R$ , then find $x \in I$ and $y \in J$ such that $1 = x + y$ . Then for any $z \in I \cap J$ , $z \in J$ implies

$z = z \cdot 1 = zx + zy = xz + zy \in I \cdot J,$

and the first isomorphism theorem for rings yields

$R/(I \cdot J) \cong R/ I \oplus R/ J.$

Given ideals $I_1,\dots, I_k \subseteq R$ , inductively define

$\displaystyle \prod_{j=1}^n I_j := \prod_{j=1}^{n-1} I_j \cdot I_n.$

Furthermore suppose that $R$ is commutative and $1 \in R$ .

Problem 6. Given ideals $I_1,\dots, I_k$ , suppose $I_m + I_n = R$ for any $m \neq n$ . Prove that

$\displaystyle R/ \prod_{j=1}^k I_j \cong \bigoplus_{j=1}^k R/I_j.$

(Click for Solution)

Solution. We proceed by induction on $k \geq 2$ . The case $k = 2$ holds by Problem 5. For the general case, define $I := \prod_{j=1}^k I_j$ . The property $I_j + I_{k+1} = R$ implies that $I + I_{k+1} = R$ and hence, so that by the $k =2$ case and the induction hypothesis,

$\begin{aligned} R/ \prod_{j=1}^{k+1} I_j &\cong R/(I \cdot I_{k+1}) \\ &\cong R/I \oplus R/I_{k+1} \\ &\cong \bigoplus_{j=1}^k R/I_j \oplus R/I_{k+1}\\ &\cong \bigoplus_{j=1}^{k+1} R/I_j.\end{aligned}$

Problem 7. Deduce the generalised Chinese remainder theorem: given integers $a_1,a_2\dots, a_k$ and $n_1,n_2,\dots, n_k$ such that

$i \neq j \quad \Rightarrow \quad \gcd(n_i, n_j) = 1,$

there exists a unique integer $0 \leq x < n_1 n_2 \cdot \dots \cdot n_k$ such that

$x \equiv_{n_i} a_i,\quad 1 \leq i \leq k.$

(Click for Solution)

Solution. It is obvious that $\mathbb Z$ is a commutative ring with $1$ . Recall that for each $j$ , $n_j \mathbb Z \subseteq \mathbb Z$ as an ideal. Given $i \neq j$ , since $\gcd(n_i, n_j) = 1$ , use Bézout’s lemma to produce integers $r, s$ such that

$n_i r + n_j s = 1.$

This result yields the property $n_i \mathbb Z + n_j \mathbb Z = \mathbb Z$ . By Problem 6,

$\displaystyle \mathbb Z/\prod_{i=1}^k n_i \mathbb Z \cong \bigoplus_{i=1}^k \mathbb Z/n_i\mathbb Z.$

Denoting $n := n_1 n_2 \cdot \dots \cdot n_k$ , Euclid’s lemma implies that

$\displaystyle \prod_{i=1}^k n_i \mathbb Z = \bigcap_{i=1}^k n_i \mathbb Z = \left( \prod_{i=1}^k n_i \right) \mathbb Z = n \mathbb Z,$

so that

$\displaystyle \mathbb Z/n \mathbb Z \cong \bigoplus_{i=1}^k \mathbb Z/n_i\mathbb Z.$

Given $a_1,\dots, a_k$ , we have

$\displaystyle (a_1 + n_1 \mathbb Z, \dots, a_k + n_k \mathbb Z) \in \bigoplus_{i=1}^k \mathbb Z/n_i\mathbb Z.$

Thanks to the isomorphism, there exists some unique $0 \leq x < n$ such that

$x + n_i \mathbb Z = a_i + n_i \mathbb Z,$

implying $x_i \equiv_{n_i} a_i$ for each $i$ .

Proof of Theorem 1. Set $k = 2$ in Problem 6.

—Joel Kindiak, 26 Jan 26, 1742H
May 1, 2026
Ring Isomorphisms
Let $R, S$ be rings.

Recall the first isomorphism theorem for rings.

Theorem 1. Let $\phi : R \to S$ be a ring homomorphism. Then $\ker(\phi)$ is an ideal and $R/{\ker(\phi)} \cong \phi(R)$ as rings.

Problem 1. Suppose $K \subseteq R$ as a sub-ring and $I \subseteq R$ as an ideal. Show that:
- $K + I \subseteq R$ as a sub-ring,
- $K \cap I \subseteq K$ as an ideal,
- $K \cap (K+I) = K$ .
Deduce that

$(K+I)/I \cong (K \cap (K+I))/(K \cap I) = K/(K \cap I).$

This result is known as the second isomorphism theorem for rings.
(Click for Solution)

Solution. We first verify the three properties.
- Sum of sub-rings remain as sub-rings.
- For the ideal claim, fix $x \in K$ and $z \in K \cap I$ . Since $I \subseteq R$ as an ideal, $xz \in I$ . Since $K$ is a sub-ring, $xz \in K$ . Therefore, $xz \in K \cap I$ . Since $I$ is also a right-ideal, we have $zx \in K \cap I$ . Therefore, $K \cap I \subseteq K$ as an ideal.
- For the final claim, the direction $(\subseteq)$ is trivial, while the direction $(\supseteq)$ holds via $x = x+ 0 \in K + I$ for any $x \in K$ .
Hence, define the map $\phi : K \to (K + I)/I$ by $x \mapsto x + I$ , which is a surjective group homomorphism under addition with kernel $K \cap I$ . Since $I \subseteq K +I$ as an ideal, we also obtain a ring structure: for $x,y \in K \subseteq K+I$ ,

$\phi(x \cdot y) = x \cdot y + I = (x+I) \cdot (y+I) = \phi(x) \cdot \phi(y).$

By the first isomorphism theorem,

$K/(K \cap I) = K/{\ker(\phi)} \cong \phi(K) = (K+I)/I.$
Problem 2. Suppose $I \subseteq J \subseteq R$ as ideals. Show that:
- $J/I \subseteq R/I$ as an ideal,
- $(R/I)/(J/I) \cong R/J$ .
This result is known as the third isomorphism theorem for rings.

(Click for Solution)

Solution. Check that the map $\phi : R/I \to R/J$ defined by $r + I \mapsto r + J$ is a surjective ring homomorphism with kernel $J/I$ and conclude using the first isomorphism theorem.

Remark 1. Let $I \subseteq R$ be an ideal and $\phi : R \to R/I$ denote the canonical quotient ring homomorphism $r \mapsto r + I$ , so that $\ker(\phi) = I$ . Define the collections

$\begin{aligned} \Sigma_1 &:= \{ K \subseteq R\ \text{sub-ring} : I \subseteq K\}, \\ \Sigma_2 &:= \{ J \subseteq R/I\ \text{sub-ring}\}. \end{aligned}$

Then the map $\psi : \Sigma_1 \to \Sigma_2$ defined by $S \mapsto S/I$ is a well-defined bijection. This result is known as the correspondence theorem for rings.

—Joel Kindiak, 22 Jan 26, 1833H
April 30, 2026
Subrings and Ideals
When discussing groups, we explored subgroups and saw how useful they help us analyse (and even connect) algebraic objects. Could we say the same for subrings?

Let $R$ be any ring, not necessarily commutative.

Definition 1. We call $S \subseteq R$ a sub-ring of $R$ if $S$ is a ring.

Lemma 1. Fix $\emptyset \neq S \subseteq R$ . Then $S$ is a sub-ring of $R$ if and only if the following conditions hold:
- For any $x,y \in S$ , $x-y \in S$ .
- For any $x,y \in S$ , $xy \in S$ .
Proof. The first criterion holds if and only if $S$ is an additive subgroup of $R$ .

$(\Rightarrow)$ If $S$ is a sub-ring of $R$ , then it must be an additive subgroup of $R$ , and the first criterion holds. It must also be closed under multiplication, so the second criterion holds.

$(\Leftarrow)$ By the first criterion, $S$ is an additive subgroup of $R$ , and is thus Abelian. By the second criterion, $S$ is closed under multiplication. Since $S$ inherits the ring properties of $R$ , we have that $S$ is a sub-ring of $R$ .

Example 1. For any $k \in \mathbb N^+$ , $k\mathbb Z$ is a sub-ring of $\mathbb Z$ .

Remark 1. If we defined rings to require the multiplicative identity, then Lemma 1 no longer holds, and sub-rings become rather useless objects. This deficiency hints at us to re-think of algebraic structures more useful than the sub-ring.

Theorem 1. Let $R, S$ be rings. A map $\phi : R \to S$ is called a ring homomorphism if for any $x,y \in R$ ,

$\phi(r + s) = \phi(r) + \phi(s),\quad \phi(r \cdot s) = \phi(r) \cdot \phi(s).$

The following hold:
- For any subring $K$ of $R$ , $\phi(K)$ is a subring of $S$ .
- For any subring $L$ of $S$ , $\phi^{-1}(L)$ is a subring of $R$ .
In particular, $\ker(\phi) := \phi^{-1}(\{0\})$ and $\phi(L)$ have ring structures. If $\phi$ is bijective, we call $\phi$ a ring isomorphism, and we say that $R, S$ are isomorphic as rings, denoted $R \cong S$ .

Proof. Since $\phi : (R, +) \to (S, +)$ is a group homomorphism, $\phi(K)$ forms an Abelian subgroup of $S$ . Closure of $\phi(K)$ under multiplication follows from $K$ being closed under multiplication and $\phi$ being a ring homomorphism. The case for $\phi^{-1}(L)$ is similar.

Recall that given a group homomorphism $\phi : (G,+) \to (H,+)$ , there exists a canonical isomorphism $\tilde \phi : (G/{\ker(\phi)}, +) \to (\phi(G), +)$ such that

$\phi = \tilde{\phi} \circ \pi_{\phi},\quad \pi_{\phi}(x) := x +\ker(\phi).$

We call this result the first isomorphism theorem for groups. Do we get a similar property for rings?

Theorem 2. Let $\phi : R \to S$ be a ring homomorphism. Then there exists a canonical ring isomorphism $\tilde \phi : R/{\ker(\phi)} \to \phi(R)$ such that

$\phi = \tilde{\phi} \circ \pi_{\phi},\quad \pi_{\phi}(x) := x +\ker(\phi).$

So yes, the first isomorphism theorem holds for rings.

Proof. Since $\phi$ is a group homomorphism under addition, the first isomorphism theorem holds for groups. In particular, $(R/{\ker(\phi)}, +)$ forms a group, and there exists a bijection $f := \tilde{\phi} : R/{\ker(\phi)} \to \phi(R)$ .

Denote $[x] \equiv x + \ker(\phi)$ for brevity. Define the ring structure on $R/{\ker(\phi)}$ using the ring structure on $\phi(R)$ : define multiplication by

$[x] \cdot [y] := f^{-1}(f([x]) \cdot f([y])).$

Then it is not hard to verify the ring structure on $R/{\ker(\phi)}$ and hence, $f$ becomes a ring homomorphism trivially. Furthermore,

$[x] \cdot [y] = [xy]$

by the following argument:

$\begin{aligned} u \in f^{-1}( f([x]) \cdot f([y]) ) \quad &\iff \quad f([u]) = f([x]) \cdot f([y]) \\ &\iff \quad \phi(u) = \phi(x \cdot y) \\ &\iff \quad u \in [x \cdot y]. \end{aligned}$

Now the group operation

$(x + \ker(\phi)) + (y + \ker(\phi)) = (x+y) + \ker(\phi)$

works because $\ker(\phi) \leq R$ forms a normal subgroup. We have defined

$(x + \ker(\phi)) \cdot (y + \ker(\phi)) := x \cdot y + \ker(\phi),$

which works by Theorem 2.

Question 1. For what other subrings $I$ of $R$ would the definition

$(x + I) \cdot (y + I) = x \cdot y + I$

work?

Lemma 2. For $I, J \subseteq R$ , interpret

$I \odot J := \{x \cdot y : x \in I, y \in J\}.$

Then

$(x + \ker(\phi)) \odot (y + \ker(\phi)) \subseteq x \cdot y + \ker(\phi).$

Proof. Given $z_1, z_2 \in \ker(\phi)$ ,

$\begin{aligned} (x+z_1) \cdot (y+z_2) &= xy + xz_2 + z_1y + z_1z_2.\end{aligned}$

The terms after $xy$ belong to $\ker(\phi)$ since

$\begin{aligned} \phi(xz_2 + z_1y + z_1z_2) &= \phi(x) \cdot \phi(z_2) + \phi(z_1) \cdot \phi(y) + \phi(z_1)\cdot \phi(z_2) \\ &= \phi(x) \cdot 0 + 0 \cdot \phi(y) + 0 \cdot 0 \\ &= 0 + 0 + 0 \\ &= 0. \end{aligned}$

Denote $I = \ker(\phi)$ in hopes of uncovering some pattern. By adopting a set-theoretic lens, we notice that for $z_1,z_2 \in I$ and $x,y \in R$ , $xz_2 \in I$ and $z_1y \in I$ .

Theorem 3. Suppose the subring $I \subseteq R$ satisfies the following properties:
- For any $x \in R$ and $z \in I$ , $xz \in I$ .
- For any $y \in R$ and $z \in I$ , $zy \in I$ .
Then for any $x,y \in R$ , $(x + I) \odot (y + I) \subseteq xy + I$ .
- If $I$ satisfies the first property, it is a left-ideal of $R$ .
- If $I$ satisfies the first property, it is a right-ideal of $R$ .
- If $I$ satisfies both properties, it is an ideal of $R$ .
Conversely, if

$(x + I) \odot (y + I) \subseteq xy + I$

for any $x,y \in R$ , then $I$ must be an ideal. As such, we can define

$(x + I) \cdot (y + I) := xy + I.$

Proof. Following the proof in Lemma 2, fix $z_1, z_2 \in I$ .
- By the first property, $xz_2 \in I$ and $z_1 z_2 \in I$ .
- By the second property, $z_1y \in I$ .
- Since $I$ is a subring of $R$ , $xz_2 + z_1y + z_1 z_2 \in I$ .
Therefore,

$\begin{aligned} (x+z_1) \cdot (y+z_2) &= xy + \underbrace{ xz_2 + z_1y + z_1z_2 }_{\in I} \in xy + I.\end{aligned}$

Conversely, setting $y = 0$ yields the left-ideal property since

$(x + I) \odot I = (x + I) \odot (0 + I) \subseteq x\cdot 0 + I = I,$

which implies that for any $x \in R$ and $z \in I$ ,

$xz = (x+0) \cdot z \in (x+I) \odot I \subseteq I.$

Similarly, setting $x = 0$ yields the right-ideal property.

Corollary 1. Let $I$ be a sub-ring of $R$ . Then $R/I$ has a ring structure if and only if $I = \ker(\phi)$ for some ring homomorphism $\phi : R \to S$ to some ring $S$ .

Proof. The direction $(\Leftarrow)$ is immediate. For the direction $(\Rightarrow)$ , Theorem 3 comes from $I$ being an ideal, so that the map $\phi : R \to R/I$ defined by $\phi(r) = r + I$ is a well-defined ring homomorphism with kernel $I$ .

The first isomorphism theorem helps us prove the Chinese remainder theorem, and we will do that as a guided exercise. For now, there are many ideal-based sub-structures worth discussing in the parlance of ring theory. We will explore these new vocabularies next time.

—Joel Kindiak, 22 Jan 26, 1834H
April 29, 2026
Law of the Rings
This ring has got nothing to do the Lord of the Rings.

The previous few posts dealt with abstract objects called groups. Really, we label a mathematical structure $G$ a group if it satisfies several properties. Then whatever is true of groups must be true of $G$ .

The integers $\mathbb Z$ were an exceedingly peculiar group under addition in the following sense:
- For any $m,n \in \mathbb Z$ , $m+n \in \mathbb Z$ .
- For any $k,m,n \in \mathbb Z$ , $k+(m+n) = (k+m) + n$ .
- $0 \in \mathbb Z$ and for any $n \in \mathbb Z$ , $0+n = n = n + 0$ .
- For any $n \in \mathbb Z$ , $-n \in \mathbb Z$ and $n+ (-n) = (-n) + n = 0$ .
Furthermore, the integers are furnished with many interesting properties:
- It is infinite, since it does not contain a finite number of elements.
- It is cyclic, since any integer $n$ can be written as a sum of $1$ or $-1$ terms.
- Its subgroups are multiples of itself (i.e. $k \mathbb Z \leq \mathbb Z$ ).
- It is Abelian under addition, that is, addition is commutative.
- Hence, all of its subgroups are normal, and $\mathbb Z_k := \mathbb Z/ k \mathbb Z$ forms a group.
- In fact, $\mathbb Z_k \cong \langle \zeta_k \rangle$ , where $\zeta_k$ is a primitive $k$ -th root of unity.
However, we have not once discussed the multiplicative properties of $\mathbb Z$ , which are just as uncanny, if not more so.

Lemma 1. The following multiplicative properties hold for $\mathbb Z$ .
- For any $m,n \in \mathbb Z$ , $m \cdot n \in \mathbb Z$ .
- For any $k,m,n \in \mathbb Z$ , $k \cdot (m \cdot n) = (k \cdot m) \cdot n$ .
- $1 \in \mathbb Z$ and for any $n \in \mathbb Z$ , $1 \cdot n = n = n \cdot 1$ .
- For any $k,m,n \in \mathbb Z$ , $k \cdot (m + n) = (k \cdot m) + (k \cdot n)$ .
- For any $k,m,n \in \mathbb Z$ , $(k + m) \cdot n = (k \cdot n) + (m \cdot n)$ .
Proof. A technical proof requires us to use the construction of the integers and, perhaps, a ton of induction.

In this case, we call $\mathbb Z$ a ring.

Definition 1. An Abelian group $(R , + )$ with additive identity $0$ (and additive inverse $-x$ of $x$ ) is a ring if it satisfies the following properties.
- For any $r,s \in R$ , $r \cdot s \in R$ .
- For any $r,s,t \in R$ , $r \cdot (s \cdot t) = (r \cdot s) \cdot t$ .
- For any $r,s,t \in R$ , $r \cdot (s + t) = (r \cdot s) + (r \cdot t)$ .
- For any $r,s,t \in R$ , $(r + s) \cdot t = (r \cdot t) + (s \cdot t)$ .
We say that the ring has unity if there exists a multiplicative identity $1 \in R$ such that for any $r \in R$ , $1 \cdot r = r = r \cdot 1$ . Furthermore, we abbreviate $r \cdot s \equiv rs$ for simplicity. Finally, by multiplying first then adding, we suppress the parentheses for brevity:

$(r \cdot s) + (r \cdot t) \equiv rs + rt.$

Henceforth, let $(R, + , \cdot)$ denote a ring. We say that $R$ is commutative if $rs = sr$ .

Example 1. $\mathbb Z$ forms a commutative ring.

When constructing the integers (and even natural numbers), $0 n = 0$ as a definition of multiplication. However, for other kinds of objects, $0 n = 0$ as a theorem.

Theorem 1. For any $r \in R$ , $0 r = 0$ .

Proof. The first claim follows from $0 + 0 = 0$ :

$\begin{aligned} (0 + 0) r &= 0 r \\ 0 r + 0r &= 0r \\ (0r + 0r) + (-0r) &= 0r + (-0r) \\ 0r + (0r + (-0r)) &= 0r + (-0r) \\ 0r + 0 &= 0 \\ 0r &= 0. \end{aligned}$

The second claim follows from the first and that $1r = r$ :

$\begin{aligned} r + (-1)r &= 1r + (-1)r \\ &= (1 +(-1))r \\ &= 0r = 0. \end{aligned}$

Corollary 1. For any $r,s \in R$ ,

$\begin{aligned} (-r) s = -rs, \quad r (-s) = -rs, \quad (-r) (-s) = rs. \end{aligned}$

Proof. The first equality arises from Theorem 1:

$rs + (-r)s = (r+(-r))s = 0 \cdot s = 0.$

Hence, $-rs = (-r)s$ .

The second equality follows similarly:

$rs + r(-s) = r(s + (-s)) = r \cdot 0 = 0,$

so that $r(-s) = -rs$ . Finally, the last equality follows from

$(-r)(-s) = -r(-s) = -(-(rs)) = rs,$

where the last result is group-theoretic since

$-x + x = 0 \quad \Rightarrow \quad -(-x) = x.$

There are many other arithmetic properties that we usually take for granted that hold for rings in general.

Before diving into ring theory, we should ask the golden question: why bother? To solve equations. More specifically, congruence equations.

Lemma 2. Fix $k, m, n \in \mathbb Z$ with $k > 0$ . Denote elements in $\mathbb Z_k$ by $[n]_k$ , i.e.

$x \in [n]_k \quad \iff \quad k \mid (x-n).$

Recall that $[m]_k + [n]_k = [m+n]_k$ by the group structure of $\mathbb Z$ under $+$ .

Then $x \in [m]_k$ and $y \in [n]_k$ implies that $xy \in [mn]_k$ .

Proof. We remark that $x \in [m]_k$ if and only if there exists an integer $x_0$ such that $x = m + kx_0$ . Therefore,

$xy =(m + k x_0) (n + k y_0) = mn + k(nx_0 + my_0 + x_0 y_0)$

implies that $xy \in [mn]_k$ .

Henceforth, define $[m]_k \cdot [n]_k := [m \cdot n]_k$ .

Theorem 2. $\mathbb Z_k$ forms a commutative ring.

Proof. Most ring properties are consequences of the ring properties in $\mathbb Z$ . Associativity, for example follows from

$\begin{aligned} [n_1]_k \cdot ([n_2]_k \cdot [n_3]_k) &= [n_1]_k \cdot [n_2 \cdot n_3]_k \\ &= [n_1 \cdot (n_2 \cdot n_3)]_k \\ &= [ (n_1 \cdot n_2) \cdot n_3 ]_k \\ &= [n_1 \cdot n_2]_k \cdot [n_3]_k \\ &= ([n_1]_k \cdot [n_2]_k) \cdot [n_3]_k. \end{aligned}$

Example 1. Evaluate $[3]_5 \cdot [2]_5$ . Hence, determine the value of $x$ such that $[2]_5 \cdot [x]_5 = [3]_5$ , where $0 \leq x < 5$ .

Proof. By a direct calculation,

$[3]_5 \cdot [2]_5 = [3 \cdot 2]_5 = [6]_5 = [1]_5.$

Therefore, we can left-multiply $[3]_5$ on both sides:

$\begin{aligned} [3]_5 \cdot ([2]_5 \cdot [x]_5) &= [3]_5 \cdot [3]_5 \\ ([3]_5 \cdot [2]_5) \cdot [x]_5 &= [3 \cdot 3]_5 \\ [1]_5 \cdot [x]_5 &= [9]_5 \\ [x]_5 &= [4]_5. \end{aligned}$

where $[n]_5 = [n + 5t]_5$ for any integer $t$ . Since the map $\{0,1,2,3,4\} \to \mathbb Z_5$ defined by $n \mapsto [n]_5$ is injective, $x = 4$ is the required solution.

Remark 1. In the language of congruence equations and modular arithmetic, we seek the set of values of $x$ such that

$5 \mid (2x-3) \quad \iff \quad 2x \equiv 3 \mod 5 \quad \iff \quad 2x \equiv_5 3.$

However, since we like to keep equality as equality, we will work in the language of congruence classes $[x]_k$ . In particular, solving congruence equations modulo $k$ amounts to solving equations in the ring $\mathbb Z_k$ .

Example 2. Show that there are no values of $x$ such that $[2]_4 \cdot [x]_4 = [3]_4$ .

Proof. Suppose for a contradiction that there exists some $0 \leq x< 4$ such that $[2]_4 \cdot [x]_4 = [2x]_4$ . Now

$\begin{aligned} [2x]_4 = [3]_4 \quad &\Rightarrow \quad 4 \mid (2x-3) \\ &\Rightarrow \quad 2 \mid (2x-3), \end{aligned}$

a blatant contradiction. Therefore, for any $x$ , $[2]_4 \cdot [x]_4 \neq [3]_4$ .

When then, can we solve congruence equations? Furthermore, can we solve systems of congruence equations?

Lemma 3. Fix integers $k, n$ with $k > 0$ . Then there exists a unique integer $0 \leq m < k$ such that

$[n]_k \cdot [m]_k = [m]_k \cdot [n]_k = [1]_k$

if and only if $\gcd(n, k) = 1$ . In this case, we denote $[m]_k =: [n]_k^{-1}$ , and we have $([n]_k^{-1})^{-1} = [n]_k$ .

Proof. By multiplication of congruence classes,

$[m \cdot n]_k = [m]_k \cdot [n]_k = [1]_k \quad \iff \quad k \mid (mn - 1).$

The latter holds if and only if there exists an integer $\ell$ such that

$mn - 1 = k\ell.$

$(\Rightarrow)$ Write

$mn - 1 = k\ell \quad \iff \quad n \cdot m + k \cdot (-\ell) = 1.$

By Bézout’s lemma, $\gcd(n, k) = 1$ .

$(\Leftarrow)$ Conversely, use Bézout’s lemma to furnish integers $r,s$ such that

$r \cdot n + s \cdot k = 1 \quad \iff \quad rn - 1 = k(-s).$

Use the division algorithm to furnish integers $q, m$ with $0 \leq m < k$ such that

$r = qk + m\quad \Rightarrow \quad mn - 1 = k(-s-qn).$

Set $\ell := -s - qn$ to yield the desired result.

Example 3. $[2]_5^{-1} = [3]_5$ and $[3]_5^{-1} = ([2]_5^{-1})^{-1} = [2]_5$ .

Proof. Since $2, 5$ are distinct prime numbers, $\gcd(2, 5) = 1$ . Write

$5 = 2 \cdot 2 + 1 \quad \Rightarrow \quad (-2) \cdot 2 - 1 = 5 \cdot (-1).$

By the division algorithm, $-2 = (-1) \cdot 5 + 3$ , so that $m = 3$ as per Lemma 3, and

$[2]_k^{-1} = [m]_k = [3]_k.$

Theorem 3. Let $a,b,k$ be integers with $k > 0$ . If $\gcd(a,k) = 1$ , then there exists a unique $0 \leq x < k$ such that

$[a]_k \cdot [x]_k = [b]_k.$

Proof. By Lemma 3, $[a]_k^{-1}$ exists, so that we left-multiply by $[a]_k^{-1}$ :

$\begin{aligned} [a]_k^{-1} \cdot ([a]_k \cdot [x]_k) &= [a]_k^{-1} \cdot [b]_k. \end{aligned}$

The left-hand side simplifies to

$\begin{aligned} [a]_k^{-1} \cdot ([a]_k \cdot [x]_k) &= ([a]_k^{-1} \cdot [a]_k) \cdot [x]_k \\ &= [1]_k \cdot [x]_k \\ &= [x]_k. \end{aligned}$

Therefore,

$\begin{aligned} [x]_k &= [a]_k^{-1} \cdot ([a]_k \cdot [x]_k) = [a]_k^{-1} \cdot [b]_k. \end{aligned}$

Denote $\mathbb Z_k^* := \mathbb Z_k \backslash \{ [0]_k\}$ . Let $p$ be a prime.

Lemma 4. $\mathbb Z_p^*$ forms a group of order $p-1$ under multiplication.

Proof. Write $\mathbb Z_p^* = \{ [1]_p, [2]_p, \dots, [p-1]_p \}$ . For each $1 \leq n \leq p-1$ , by Lemma 3,

$\begin{aligned} p \nmid n \quad &\Rightarrow \quad \gcd(n, p) = 1 \\ &\Rightarrow \quad[n]_p^{-1} \in \mathbb Z_p^*. \end{aligned}$

Theorem 4 (Fermat’s Little Theorem). For any $1 \leq a \leq p-1$ ,

$[a]_p^{p-1} = [1]_p.$

More generally, for any integer $a$ , $[a]_p^p = [a]_p$ .

Proof. By Lemma 4, $\mathbb Z_p^*$ forms a group of order $p-1$ under multiplication. Furthermore, $\langle [a]_p \rangle \leq \mathbb Z_p^*$ . By Lagrange’s theorem

$m:= | \langle [a]_p \rangle | \mid | \mathbb Z_p^* | = p-1.$

Find an integer $k$ such that $p-1 = k m$ . Then

$\begin{aligned} [a]_p^{p-1} &= [a]_p^{k m} \\ &= ([a]_p^{ m })^k \\ &= [1]_p^k = [1]_p. \end{aligned}$

The result holds trivially for $a = 0$ . For the general case, use the division algorithm to obtain unique integers $q, r$ such that

$a = qp + r,\quad 0 \leq r < p.$

Then the special case yields

$[a]_p^{p-1} = [r]_p^{p-1} = [1]_p.$

Definition 2. A ring $R$ is a field if $R^* := R\backslash \{0\}$ forms an Abelian group under multiplication.

Example 4. For prime numbers $p$ , $\mathbb Z_p$ forms a finite field with $p$ elements, and is often denoted $\mathbb F_p \equiv \mathbb Z_p$ to emphasise its field structure.

Example 5. The commonly used infinite sets $\mathbb Q, \mathbb R, \mathbb C$ form fields.

Now we could go down the number-theoretic rabbit-hole a bit further, but I hope I’ve given sufficient examples in number theory that motivate a study of ring theory. Next time, we will look at the various sub-structures, and in particular, ideals.

—Joel Kindiak, 20 Jan 26, 1950H
April 28, 2026
Euler’s Coprimes

Problem 1. Given integers $a, b, k$ , show that $\gcd(a,b) = \gcd(a+kb, b)$ .

(Click for Solution)

Solution. Let $c = \gcd(a,b)$ and $d = \gcd(a+kb, b)$ . Then

$c \mid a\quad \text{and} \quad c \mid b \quad \Rightarrow \quad c \mid (a+kb),$

so that $c \mid d$ . Similarly,

$d \mid (a+kb)\quad \text{and} \quad d \mid b \quad \Rightarrow \quad d \mid ((a+kb) - kb)= a,$

so that $d \mid c$ .

Let $n > 1$ denote any positive integer.

Definition 1. Define

$\Phi(n) := \{ 1 \leq x \leq n : \gcd(x, n) = 1\}$

and Euler’s totient function $\varphi$ by $\varphi(n) := |\Phi(n)|$ .

Problem 2. Show that

$G_n := \{ [x]_n : x \in \Phi(n)\} \subseteq \mathbb Z_n^*$

forms an Abelian group of order $\varphi(n)$ under multiplication.

(Click for Solution)

Solution. To establish closure, we need to show that for $x, y \in \Phi(n)$ , $[xy]_n = [m]_n$ for some $m \in \Phi(n)$ . To that end, use Bézout’s lemma to furnish integers $r_1, s_1, r_2, s_2$ such that

$\begin{aligned} r_1 \cdot x + s_1 \cdot n &= 1,\\ r_2 \cdot y + s_2 \cdot n &= 1. \end{aligned}$

Multiplying both equations together,

$(r_1 r_2) \cdot xy + (r_1 s_2 x + r_2 s_1 y + s_1 s_2 n) \cdot n = 1.$

By Bézout’s lemma again, $\gcd(xy, n) = 1$ .

The identity is clearly $[1]_n$ and the inverse of $[x]_n$ exists from $\gcd(x, n) = 1$ .

Problem 3 (Euler’s Theorem). Show that for any integer $a$ such that $\gcd(a, n) = 1$ ,

$[a]_n^{\varphi(n)} = [1]_n.$

(Click for Solution)

Solution. Assume $1 \leq a \leq n$ for simplicity, so that $a \in \Phi(n)$ . If $a \neq 1$ , then $\langle [a]_n \rangle \leq G_n$ . By Lagrange’s theorem,

$m := | \langle [a]_n \rangle | \mid |G_n| = \varphi(n).$

Find an integer $k$ such that $\varphi(n) = km$ . Then

$\begin{aligned} [a]_n^{\varphi(n)} &= [a]_n^{km} \\ &= ([a]_n^m)^k \\ &= [1]_n^k = [1]_n. \end{aligned}$

For the general case, use the division algorithm to find integers $q,r$ with $0 \leq r < n$ such that $a = qn + r$ . By Problem 1,

$\begin{aligned} \gcd(r,n) &= \gcd(a-qn,n) \\ &= \gcd(a,n) \\ &= 1 . \end{aligned}$

By the special case, $[a]_n^{\varphi(n)} = [r]_n^{\varphi(n)} = [1]_n$ .

Problem 4. Verify Fermat’s little theorem: for any integer $a$ and prime $p$ ,

$[a]_p^p = [a]_p.$

(Click for Solution)

Solution. The result for $a = 0$ is trivial. For $1 \leq a \leq p-1$ , $\varphi(p) = p-1$ , so that

$\begin{aligned} [a]_p^p &= [a]_p \cdot [a]_p^{p-1} \\ &= [a]_p \cdot [a]_p^{\varphi(p)} \\ &= [a]_p \cdot [1]_p \\ &= [a]_p. \end{aligned}$

For the general case, write $a = qp + r$ where $0 \leq r \leq p-1$ :

$[a]_p^p = [r]_p^p = [r]_p = [a]_p.$

—Joel Kindiak, 20 Jan 26, 1940H

April 27, 2026
The Language of Symmetry

In many ways, group theory gives language to symmetries. And the most symmetric idea we have yet is, uncreatively, the symmetric group.

Definition 1. For any set $K$ , define the symmetric group on $K$ by

$\mathrm{Sym}(K) := \{f \in \mathcal F(K, K) : f\ \text{is bijective}\}.$

In the context of linear algebra, for any finite-dimensional vector space $V$ ,

$\mathrm{Sym}(V) \cong \mathrm{Sym}(\mathbb R^{\dim(V)}).$

Theorem 1. For any set $K$ , $\mathrm{Sym}(K)$ forms a group under function composition.

Proof. For any $f, g \in \mathrm{Sym}(K)$ , $g \circ f : K \to K$ is bijective, so that $g \circ f \in \mathrm{Sym}(K)$ , so that composition yields a well-defined binary operation. Associativity holds for function composition, the identity element is given by $\mathrm{id}_K$ , and the inverse of $f$ is the inverse function $f^{-1}$ .

For any $n \geq 1$ , define $\mathcal S_n := \mathrm{Sym}(\{1,\dots, n\})$ .

Lemma 1. Given distinct $x_i$ , $\mathrm{Sym}(\{x_1,\dots, x_n\}) \cong \mathcal S_n$ .

Proof. For any $f \in \mathcal S_n$ , define $\Phi_f \in \mathrm{Sym}(\{x_1,\dots, x_n\})$ by

$\Phi_f(x_i) = x_{f(i)}.$

We claim that the map $\Phi : \mathcal S_n \to \mathrm{Sym}(\{x_1,\dots, x_n\})$ defined by $\Phi(f) = \Phi_f$ is bijective.

For injectivity, suppose $\Phi(f_1) = \Phi(f_2)$ . Then for any $x_i$ ,

$x_{f_1(i)} = \Phi_{f_1}(x_i) = \Phi_{f_2}(x_i) = x_{f_2(i)}.$

Since the $x_j$ terms are distinct, $f_1(i) = f_2(i)$ . Since this equation holds for $i = 1,\dots, n$ , we have $f_1 = f_2$ , as required.

For surjectivity, fix $g \in \mathrm{Sym}(\{x_1,\dots, x_n\})$ . Since $g$ is bijective, for each $x_i$ , there exists a unique $x_{i'}$ such that $g(x_i) = x_{i'}$ . Define $f \in \mathcal S_n$ by $f(i) = i'$ . We claim that $\Phi_f = g$ , which holds since for any $x_i$ ,

$\displaystyle \Phi_f (x_i) = x_{f(i)} = x_{i'} = g(x_i),$

so that $\Phi(f) = \Phi_f = g$ , as required.

Hence, to understand the symmetries on any finite set, it suffices to understand the symmetries on $\mathcal S_n$ .

For simplicity, we now zoom in on $\mathcal S_n$ . Each element $f \in \mathcal S_n$ is a bijection, and thus, corresponds to a permutation of the ordered $n$ -tuple $(1,\dots, n)$ . Thanks to ideas in combinatorics, there are $n!$ possible bijections, so that $|\mathcal S_n| = n!$ .

Lemma 2. For any $f \in \mathcal S_n$ , define $N$ -fold composition by

$f^N := \underbrace{f \circ \cdots \circ f}_N.$

Then for any $k = 1,\dots, n$ , there exists a unique $1 \leq N_k \leq n$ such that $f^{N_k}(k) = k$ and for any $0 < i < N_k$ , $f^i(k) \neq k$ .

Proof. Fix $k = 1,\dots, n$ . Define

$\langle k \rangle := \{ f^N(k) : N \in \mathbb N^+\} \subseteq \{1,\dots, n\}.$

Consider the set

$\{ k, f(k), \dots, f^{n-1}(k), f^n(k) \} \subseteq \{1,\dots, n\}.$

Since the left-hand side has at most $n$ elements, we must have some $i < j$ such that

$f^i(k) = f^j(k) \quad\Rightarrow \quad f^{i-j}(k) = k.$

Therefore, the set

$\{N \in \mathbb N^+ : f^N(k) = k\}$

contains some $j-i \leq n$ and thus is nonempty. By the well-ordering principle, it contains a unique least integer $N_k$ , so that $f^{N_k}(k) = k$ .

In what follows, fix a non-identity permutation $f \in \mathcal S_n$ .

Lemma 3. The relation $\sim_f$ on $\mathcal S_n$ defined by

$i \sim_f j \quad \iff \quad \exists k \in \mathbb N^+: f^k(i) = j$

is an equivalence relation on $\{1,\dots, n\}$ . Furthermore, if $i \sim_f j$ , then $N_i = N_j$ as per Lemma 2.

Proof. We need to check that $\sim_f$ is reflexive, symmetric, and transitive.

For reflexivity, for any $i$ , Lemma 2 furnishes some $N_i \in \mathbb N^+$ such that $f^{N_i}(i) = i$ .

For symmetry, suppose $i_1 \sim_f i_2$ , so that there exists $k \in \mathbb N^+$ such that $f^k(i) = i_2$ . Find $m > 0$ such that $mN_{i_2} > k$ . Then

$f^k(i_1) = i_2 = f^{mN_{i_2}}(i_2) \quad \Rightarrow \quad f^{mN_{i_2} - k}(i_2)= i_1.$

Hence, $i_2 \sim_f i_1$ . Finally for transitivity, if $i_1 \sim_f i_2$ and $i_2 \sim_f i_3$ , then there exists $k_1, k_2$ such that

$f^{k_1}(i_1) = i_2, \quad f^{k_2}(i_2) = i_3.$

By substituting, $f^{k_1+k_2}(i_1) = f^{k_2}(f^{k_1}(i_1)) = i_3$ . Hence, $i_1 \sim_f i_3$ , as required.

Finally, suppose $i \sim_f j$ for $i \neq j$ . Then there exist some smallest $k \in \mathbb N^+$ such that

$f^{k}(i) = j.$

Since $f^k(i) \neq j$ , we can assume $k < N_i$ . By definition,

$f^{N_i - k}(j) = f^{N_i-k}(f^{k}(i)) = f^{N_i}(i) = i.$

By back-substituting,

$f^{N_i}(j) = f^k(f^{N_i - k}(j)) = f^k(i) = j.$

By minimality, $N_j \leq N_i$ . By a symmetric argument, $N_i \leq N_j$ . Therefore, $N_i = N_j$ , as required.

Lemma 4. Using $\sim_f$ Lemma 3, find elements $i_1,\dots, i_m$ such that

$\displaystyle \{1,\dots, n\} = \bigsqcup_{k=1}^m [i_k].$

Then for each $k$ , $[i_k] = \{f^N(i_k) : N \in \mathbb N\}$ . In this case, we write

$\displaystyle f = \prod_{k=1}^m (i_k\ f(i_k)\ \cdots\ f^{N_{i_k}-1}(i_k)).$

Each cycle is an $N_{i_k}$ -cycle.

Proof. Exercise.

In particular, cyclic permutations within each cycle do not change the result: for instance,

$(1\ 2\ 3) = (2\ 3\ 1) = (3\ 1\ 2).$

As such, we can now properly define a special kind of permutation: a cycle.

Example 1. Consider the permutation $f \in \mathcal S_4$ defined by

$(f(1), f(2), f(3), f(4)) = (2, 1, 4, 3).$

Evaluate $f$ as a product of cycles.

Solution. We have $f(1) = 2$ ,

$f^2(1) = f(f(1)) = f(2) = 1,$

$f(3) = 4$ , and

$f^2(3) = f(f(3)) = f(4) = 3.$

Hence, $f = (1\ 2)(3\ 4) = (3\ 4)(1\ 2)$ .

Definition 2. A cycle is a permutation in $\mathcal S_n$ of the form $(i_1\ i_2\ \cdots\ i_m)$ for $m$ distinct integers in $\{1,\dots, n\}$ . Given the cycles

$f = (i_1\ i_2\ \cdots\ i_m),\quad g = (j_1\ j_2\ \cdots\ j_n),$

we denote the composition permutation by the product

$g \circ f = (j_1\ j_2\ \cdots\ j_n)(i_1\ i_2\ \cdots\ i_m).$

Lemma 5. Every non-identity permutation in $\mathcal S_n$ is a product of unique, disjoint cycles in $\mathcal S_n$ .

Proof. Since composition of disjoint cycles is commutative, applying Lemma 4 gives the desired result.

Can we decompose cycles further? Surprisingly, yes.

Definition 3. A transposition is a cycle in $\mathcal S_n$ of the form $(i\ j)$ . It is clear that $(i\ j)(i\ j) = \mathrm{id}$ , so that $(i\ j)^{-1} = (i\ j)$ is also a transposition.

Example 2. Evaluate the permutations

$(1\ 3)(1\ 2),\quad (1\ 2)(1\ 3),\quad (2\ 3)(1\ 2),\quad (3\ 4)(1\ 2).$

giving your answers as products of disjoint cycles.

Solution. Denote $f = (1\ 2)$ and $g =(1\ 3)$ . Then

$\begin{aligned} (g \circ f)(1) &= g(f(1)) = g(2) = 2 ,\\ (g \circ f)(2) &= g(f(2)) = g(1) = 3, \\ (g \circ f)(3) &= g(f(3)) = g(3) = 1. \end{aligned}$

Furthermore, $(1\ (g\circ f)(1)\ (g \circ f)^2(1)) = (1\ 2\ 3)$ . Therefore,

$(1\ 3)(1\ 2) = (1\ 2\ 3).$

We leave it as an exercise to check that

$(1\ 2)(1\ 3) = (1\ 3\ 2) = (2\ 3)(1\ 2).$

Finally, since $(3\ 4)(1\ 2)$ is a product of unique cycles, there is nothing to simplify.

Theorem 2. Every non-identity permutation in $\mathcal S_n$ is a product of transpositions in $\mathcal S_n$ .

Proof. Let $f \in \mathcal S_n$ be a permutation. By Lemma 5, $f$ can be written as a product of (disjoint) cycles. Hence, it suffices to prove the case for a non-trivial cycle $f$ . Without loss of generality, assume

$f = (1\ 2\ \cdots \ k),$

with the general case obtained via relabeling. We will prove by induction on $k$ , where $2 \leq k \leq n$ . The case $k = 2$ is obvious: $(1\ 2)$ is itself already a transposition. For the general case, suppose the cycle

$f := (1\ 2\ \cdots \ i).$

can be written as a product of transpositions. Note that $f(i+1) = i+1$ . Consider the cycle

$g := (1\ 2\ \cdots \ i \ i+1)$

of length $i+1$ . Define the transposition $\phi := (1, i+1)$ . We observe that

$\begin{aligned} (\phi \circ f)(i) &= \phi(f(i)) = \phi(1) = i+1 = g(i) \end{aligned}$

and

$\begin{aligned} (\phi \circ f)(i+1) &= \phi(f(i+1)) = \phi(i+1) = 1 = g(i+1). \end{aligned}$

Furthermore, for $1 \leq k \leq i-1$ , $g(k) = k+1$ , so that

$( \phi \circ f )(k) = \phi(f(k)) = \phi(k+1) = k+1 = g(k).$

Therefore, $g = \phi \circ f$ , which means

$g = (1\ i+1)f$

is a product of transpositions, as required.

Strictly speaking, any permutation can be lengthened arbitrarily by composing with $(1\ 2)(1\ 2) = \mathrm{id}$ on either side. Nonetheless, the number of transpositions either remains even or remains odd, which motivates the next definition.

Definition 4. Call a permutation even (resp. odd) if it is a product of an even number (resp. odd number) of transpositions. Define the alternating group $\mathcal A_n$ by

$\mathcal A_n := \{f \in \mathcal S_n : f\ \text{is even}\}.$

Lemma 6. $\mathcal A_n \triangleleft \mathcal S_n$ . Furthermore, $|\mathcal A_n| = |\mathcal S_n|/2$ .

Proof. Fix $f \in \mathcal A_n$ and $g = \phi_n \circ \cdots \circ \phi_1 \in \mathcal S_n$ for transpositions $\phi_1,\dots, \phi_n$ . It is clear that

$g^{-1} = \phi_1^{-1} \circ \cdots \circ \phi_n^{-1} \in \mathcal S_n$

and each $\phi_i^{-1}$ is a transposition. Suppose $f$ is a product of $m$ transpositions for some even number $m$ . Then $g^{-1} \circ f \circ g$ is a product of $m + 2n$ transpositions. Since $m+2n$ is even, $g^{-1} \circ f \circ g \in \mathcal A_n$ . Hence, $\mathcal A_n \triangleleft \mathcal S_n$ .

For the counting claim, we note that $\mathcal A_n \sqcup (1\ 2)\mathcal A_n = \mathcal S_n$ , and that there exists an obvious bijection $\mathcal A_n \to (1\ 2)\mathcal A_n$ given by $f \mapsto (1\ 2)f$ . Hence,

$\begin{aligned} |\mathcal S_n| &= |\mathcal A_n \sqcup (1\ 2)\mathcal A_n| \\ &= |\mathcal A_n| + |(1\ 2)\mathcal A_n| \\ &= |\mathcal A_n| + |\mathcal A_n| = 2 |\mathcal A_n|,\end{aligned}$

so that $|\mathcal A_n| = |\mathcal S_n|/2$ .

Lemma 7. Given $\mathcal K \triangleleft \mathcal A_n$ with $\mathcal K \neq \{\mathrm{id}\}$ , if $\mathcal K$ contains a $3$ -cycle, then $\mathcal K = \mathcal A_n$ .

Proof. Suppose $\mathcal K \triangleleft \mathcal A_n$ and $\mathcal K \neq \{\mathrm{id}\}$ so that $\mathcal K$ contains some non-identity $3$ -cycle $f = (1\ 2\ 3) \in \mathcal A_n$ without loss of generality.

We claim that $\mathcal K$ contains all $3$ -cycles. Indeed, let $g = (i_1\ i_2\ i_3)$ be a $3$ -cycle. Define $h \in \mathcal S_n$ by

$h(k) = i_k,\quad k=1,\dots,n.$

If $h$ is even, then $h \in \mathcal A_n$ , and

$g = h \circ f \circ h^{-1} \in \mathcal K.$

If $h$ is odd, then repeat the argument by replacing $h$ with $(i_4\ i_5)h$ . Thus, $\mathcal K$ contains all $3$ -cycles.

Fix $f \in \mathcal A_n$ . We claim that $f$ can be written as a product of $3$ -cycles, so that $f \in \mathcal K$ .

We observe that without loss of generality,

$\begin{aligned} (1\ 2)(3\ 4) &= (1\ 3\ 2)(1\ 3\ 4) \in \mathcal K, \\ (1\ 2)(1\ 3) &= (1\ 3\ 2) \in \mathcal K, \\ (1\ 2)(1\ 2) &= \mathrm{id} \in \mathcal K. \end{aligned}$

Therefore, since $f$ can be written as a product of $k$ terms of the form $(a\ b)(c\ d)$ , we have $f \in \mathcal K$ . Therefore, $\mathcal K = \mathcal A_n$ .

Definition 5. Given a group $G$ and $H \leq G$ , call $H$ simple if for any $K \triangleleft H$ , $K = \{e\}$ or $K = H$ .

Remark 1. The definition of a simple group resembles that of a prime number: a positive integer $p$ is prime if it is not composite. That is, if $d \mid p$ , then $d = 1$ or $d = p$ .

Lemma 8. $\mathcal A_5$ is simple.

Proof. Fix $\mathcal K \triangleleft \mathcal A_5$ with $\mathcal K \neq \{\mathrm{id}\}$ and a non-identity permutation $f \in \mathcal K$ . By Lemma 7 it suffices to check that $\mathcal K$ has a $3$ -cycle.

Write $f$ as a product of disjoint cycles or identity maps

$f = f_k \circ \cdots \circ f_1, \quad k \leq 5.$

If all $f_i$ are the identity map, then $f$ is the identity map, a contradiction. Hence, suppose $f_1$ is not an identity map. If $f$ has no $3$ -cycles, then either $f = \phi_1$ , or $f = \phi_2 \circ \phi_1$ for some disjoint transpositions $\phi_1, \phi_2$ , or $f$ is a $4$ -cycle or $f$ is a $5$ -cycle.

In the first case, suppose without loss of generality that $f = (1\ 2)$ . Since $\mathcal K$ is normal,

$(2\ 4) = (4\ 3\ 1)(1\ 2)(1\ 3\ 4) \in \mathcal K.$

Hence, $(1\ 4\ 2) = (2\ 4) (1\ 2) \in \mathcal K$ , and $\mathcal K$ contains a $3$ -cycle.

In the second case, suppose without loss of generality that $f = (1\ 2)(3\ 4)$ . Since $\mathcal K$ is normal,

$(2\ 5)(3\ 4) = (1\ 5)(1\ 2)(3\ 4)(1\ 5) \in \mathcal K.$

Hence,

$(1\ 5\ 2) = (2\ 5)(3\ 4)(1\ 2)(3\ 4) \in \mathcal K,$

and $\mathcal K$ contains a $3$ -cycle.

In the third case, suppose $f = (1\ 2\ 3\ 4)$ . Write

$(1\ 2\ 3\ 4) = (1\ 4)(1\ 3)(1\ 2) \notin \mathcal A_5,$

so the third case is impossible.

In the fourth case, suppose $f = (1\ 2\ 3\ 4\ 5)$ . Define $g = (1\ 2)(3\ 4)$ . Then

$(2\ 5\ 4) = g \circ f \circ g^{-1} \circ f \in \mathcal K.$

Hence, no matter which case, $\mathcal K$ contains a $3$ -cycle.

By Lemma 7, $\mathcal K = \mathcal A_5$ . Hence, $\mathcal A_5$ is simple.

Lemma 9. $\mathcal A_6$ is simple.

Proof. Fix $\mathcal K \triangleleft \mathcal A_6$ with $\mathcal K \neq \{ \mathrm{id} \}$ . Let $f \in \mathcal K$ be a non-identity even permutation. By Lemma 7 it suffices to show that $\mathcal K$ contains a $3$ -cycle. Define the subset

$\mathcal G := \{g \in \mathcal K : g(6)= 6\} \subseteq \mathcal K.$

We claim that $\mathcal G \backslash \{\mathrm{id}\} \neq \emptyset$ . If $f(6) = 6$ , then we are done. Otherwise, after relabelling, $f = (1\ 2\ 3\ 4)(5\ 6)$ or $f = (1\ 2\ 3)(4\ 5\ 6)$ .

In the first case, $g := f^2 = (1\ 3)(2\ 4) \in \mathcal K$ and hence $g(6)= 6$ , as required. In the second case, define $h = (2\ 3\ 4) \in \mathcal A_6$ . Since $\mathcal K \triangleleft \mathcal A_6$ ,

$g := f\circ (h \circ f^{-1} \circ h^{-1}) \in \mathcal K$

and $g(6) = 6$ , as required. Therefore, the sub-group $\mathcal G := \{g \in \mathcal K : g(6)= 6\} \leq \mathcal K$ contains a non-identity element.

Identify $\mathcal A_{n-1} \cong \{ g \in \mathcal A_n : g(n) = n \} \leq \mathcal A_6$ so that $\{\mathrm{id}\} \neq \mathcal G \cap \mathcal A_5 \triangleleft \mathcal A_5$ .

By Lemma 8, $\mathcal A_5$ being simple implies that $\mathcal G \cap \mathcal A_5 = \mathcal A_5$ , so that $\mathcal A_5 \subseteq \mathcal G \subseteq \mathcal K$ .

Since $\mathcal A_5$ contains a $3$ -cycle, so does $\mathcal K$ . By Lemma 7, $\mathcal K = \mathcal A_6$ , so that $\mathcal A_6$ is simple, as required.

Theorem 3. $\mathcal A_n$ is simple for $n \geq 5$ .

Proof. By Lemmas 8 and 9, it suffices to prove that $\mathcal A_n$ is simple for $n \geq 7$ . Fix $\{\mathrm{id}\} \neq \mathcal K \triangleleft \mathcal A_n$ . Then there exists $f \in \mathcal K$ such that $f \neq \mathrm{id}$ .

Without loss of generality, suppose $f(1) = 2$ . Define $g := (2\ 3\ 4) \in \mathcal A_n$ . Define $h := f^{-1} \circ (g \circ f \circ g^{-1}) \in \mathcal K$ . Then $h(1) = f^{-1}(3) \neq 1$ , so $h \neq \mathrm{id}$ . Furthermore, if $f(i) \notin \{2, 3, 4\}$ , then

$(f^{-1} \circ g \circ f)(i)= i.$

Thus, $f^{-1} \circ g \circ f = (f^{-1}(2)\ f^{-1}(3)\ f^{-1}(4))$ and $g = (2\ 3\ 4)$ are $3$ -cycles. Hence, $h$ permutes at most $6$ elements. Suppose for simplicity that $h$ permutes $\{1,2,3,4,5,6\}$ . Since $n \geq 7$ , $h(n) = n$ .

Following the argument in Lemma 9, $\mathcal A_5 \subseteq \mathcal A_6 \subseteq \mathcal K$ . Since $\mathcal A_5$ contains a $3$ -cycle, so does $\mathcal K$ . By Lemma 7, $\mathcal K = \mathcal A_n$ . Hence, $\mathcal A_n$ is simple.

Theorem 3 turns out to be a key ingredient in establishing the non-existence of the quintic formula. But to get there, we will need to develop a lot more abstract algebra machinery.

For now, we conclude our discussions on $\mathcal S_n$ by making reference to Cayley’s theorem.

Theorem 4 (Cayley). If $G$ is a finite group, then there exists $H \leq \mathcal S_{|G|}$ such that $G \cong H$ .

Proof. For each $g \in G$ , define $\Phi_g \in \mathrm{Sym}(G)$ by $\Phi_g(x) = gx$ . Define the map $\Phi : G \to \mathrm{Sym}(G)$ by $\Phi(g) = \phi_g$ . We claim that $\Phi$ is an injective homomorphism. For injectivity, suppose $\Phi(g_1) = \mathrm{id}$ so that $\Phi_{g_1} = \mathrm{id}$ . Then

$g_1 = g_1 e = \mathrm{id}(e) = e,$

as required. For the homomorphism property, fix $x \in G$ . Then

$\Phi_{g_1 g_2}(x) = (g_1 g_2)x = g_1(g_2 x) =g_1 \Phi_{g_2}(x) = (\Phi_{g_1} \circ \Phi_{g_2})(x).$

Hence, $\Phi(g_1 g_2) = \Phi_{g_1 g_2} = \Phi_{g_1} \circ \Phi_{g_2}$ , as required. By Lemma 1,

$G \cong \Phi(G) \leq \mathrm{Sym}(G) \cong \mathcal S_{|G|}.$

Letting $f : \mathrm{Sym}(G) \to \mathcal S_{|G|}$ denote a group isomorphism, define $H := (f \circ \Phi)(G)$ so that

$G \cong \Phi(G) \cong (f \circ \Phi)(G) = H.$

Remark 2. If $G$ is infinite, we still have the relationship

$G \cong \Phi(G) \leq \mathrm{Sym}(G).$

This means that any study of finite groups, at its core, reduces to a study of $\mathcal S_n$ .

Next time, we re-visit an important finite group $\mathbb Z_k = \mathbb Z/k\mathbb Z$ , and consider multiplication, venturing our toes into an extension of groups—rings.

—Joel Kindiak, 31 Dec 25, 2210H

April 24, 2026
Group Isomorphisms

We have previously seen that $\mathbb Z/k\mathbb Z \cong \langle \zeta_k \rangle$ .

Denoting the left-hand side by $\mathbb Z_k$ , we have $\mathbb (\mathbb Z_k, +) \cong (\langle \zeta_k \rangle, \cdot)$ .

We want to generalise this observation into a unified criterion that determines groups that are isomorphic to each other. In what follows, let $G$ be a group.

Definition 1. Given $K, M \subseteq G$ , define

$\displaystyle MK := \{mk : k \in K, m \in M\} \subseteq G.$

and $gK := \{g\}K$ . Clearly, $eK = K$ . Define $G/K := \{gK : g \in G\}$ .

Lemma 1. Suppose that $K \leq G$ . We have

$\forall g_1, g_2 \in G\quad (g_1 K) (g_2 K) \supseteq (g_1 g_2) K.$

Furthermore,

$\forall g_1, g_2 \in G\quad (g_1 K) (g_2 K) = (g_1 g_2) K$

if and only if

$\forall g \in G\quad \forall k \in K\quad gkg^{-1} \in K.$

In this case, we say that $K$ is normal, and denote $K \triangleleft G$ .

Proof. The first claim holds since for any $k \in K$ , $e \in K$ implies that

$(g_1 g_2)k = \underbrace{ (g_1 e) }_{\in g_1 K} \underbrace{ (g_2 k) }_{\in g_2 K} \in (g_1 K)(g_2 K).$

For the biconditional, we prove in both directions.

$(\Rightarrow)$ Fix $g \in G$ and $k \in K$ . Since $K \leq G$ , $e \in K$ . Therefore,

$\begin{aligned} gkg^{-1} &= \underbrace{ (gk) }_{\in gK} \underbrace{ (g^{-1} e) }_{\in g^{-1}K} \\ &\in (gK)(g^{-1}K) \\ &= (gg^{-1})K = eK = K. \end{aligned}$

$(\Leftarrow)$ Fix $g_1, g_2 \in G$ . It suffices to prove

$(g_1 K) (g_2 K) \subseteq (g_1 g_2) K.$

Indeed, inclusion holds since for any $k_1, k_2 \in K$ ,

$\begin{aligned} (g_1 k_1) (g_2 k_2) &= g_1 e k_1 g_2 k_2 \\ &= g_1 (g_2 g_2^{-1})k_1 g_2 k_2 \\ &= (g_1 g_2) \underbrace{ \underbrace{(g_2^{-1} k_1 g_2)}_{\in K} \underbrace{ k_2 }_{\in K} }_{\in K} \in (g_1 g_2) K. \end{aligned}$

Lemma 1 implies that we can define a group structure on $G/K$ if and only if $K$ is a normal subgroup of $G$ . The next lemma tells us that the only real subgroups that are normal are kernels of homomorphisms.

Lemma 2. $K \triangleleft G$ if and only if there exists a group $H$ and a surjective homomorphism $\phi : G \to H$ such that $K = \ker(\phi)$ .

Proof. $(\Leftarrow)$ It suffices to check that $\ker(\phi) \triangleleft G$ , which holds since for any $g \in G$ and $k \in \ker(\phi)$ ,

$\begin{aligned} \phi(gkg^{-1}) &= \phi(g)\phi(k)\phi(g^{-1}) \\ &= \phi(g) e \phi(g)^{-1} \\ &= \phi(g) \phi(g)^{-1} = e. \end{aligned}$

Hence, $gkg^{-1} \in \ker(\phi)$ , so that $\ker(\phi) \triangleleft G$ , as required.

$(\Rightarrow)$ Recall the equivalence relation $\sim_K$ on $G$ defined by

$g_1 \sim_K g_2 \quad \iff \quad g_1 g_2^{-1} \in K.$

Using discrete mathematics, the projection map $\pi_K : G \to G/{\sim_K}$ defined by

$\pi_K(g) = [g]$

is a well-defined surjection. It is also clear that the computation

$g' \sim_K g \quad \iff \quad g' g^{-1} \in K \quad \iff \quad g' \in gK$

implies that $G/{\sim_K} = G/K$ . By Lemma 1, $G/K$ has a well-defined binary operation given by

$(g_1 K) (g_2 K) = (g_1 g_2) K,\quad g_1,g_2 \in G,$

and it is straightforward to verify that $G/K$ forms a group with identity $eK = K$ under this operation.

Now, define $H := G/K$ and $\phi := \pi_K : G \to H$ be defined as above. It remains to check that $\ker(\phi) = K$ :

$\begin{aligned} u \in \ker(\phi) \quad &\iff \quad \phi(u) = K \\ &\iff \quad \pi_K(u) = K \\ &\iff \quad [u] = K \\ &\iff \quad u \in K, \end{aligned}$

as required.

It is the normal subgroup property of the kernel that underscores its ubiquity in abstract algebra. In particular, it plays a crucial role in determining when two groups are isomorphic to each other.

Theorem 1. For any group homomorphism $\phi : G \to H$ , $G/{\ker(\phi)} \cong \phi(G)$ . This result is known as the first isomorphism theorem.

Proof. For simplicity, first suppose $\phi$ is surjective so that $\phi(G) = H$ . By Lemma 2, $\ker(\phi) \triangleleft G$ . Using discrete mathematics, the projection map $\pi_{\ker(\phi)}$ induces a bijection $\tilde \phi : G/{\ker(\phi)} \to H$ such that $\phi = \tilde \phi \circ \pi_{\ker(\phi)}$ .

It remains to check that $\tilde \phi : G/{\ker(\phi)} \to H$ is a group homomorphism. Indeed, since $\pi_{\ker(\phi)} : G \to G/{\ker(\phi)}$ is surjective, for any $g \in G$ ,

$\begin{aligned} \tilde\phi([g]) &= \tilde \phi(\pi_{\ker(\phi)}(g)) \\ &= (\tilde \phi \circ \pi_{\ker(\phi)})(g) = \phi(g). \end{aligned}$

Since $\phi$ is a group homomorphism, for any $g_1, g_2 \in G$ ,

$\begin{aligned} \tilde \phi([g_1][g_2]) &= \tilde \phi([g_1 g_2]) \\ &= \phi(g_1 g_2) \\ &= \phi(g_1) \phi(g_2) \\ &= \tilde \phi([g_1]) \tilde \phi([g_2]), \end{aligned}$

so that $\tilde \phi$ is a group homomorphism, as required. For the general case, it is clear that $\phi : G \to \phi(G)$ is surjective, so we recover the full result, as required.

In particular, $G \cong G/\{e\} \cong H$ if and only if $\ker(\phi) = \{e\}$ , which holds if and only if $\phi$ is injective.

Corollary 1. Given $K \triangleleft G$ and a group $H$ , if $G/K \cong H$ , then there exists a surjective homomorphism $\phi : G \to H$ .

Proof. Let $\tilde \phi : G/K \to H$ be a group isomorphism. Since $\pi_K : G \to G/K$ is surjective, simply define $\phi = \tilde \phi \circ \pi_K$ .

Example 1. $\mathbb Z/ k \mathbb Z \cong \langle \zeta_k \rangle$ .

Proof. Define the surjective homomorphism $\phi : \mathbb Z \to \mathbb C^{\times}$ by $\phi(n) = \zeta_k^n$ . Then $\ker(\phi) = k\mathbb Z$ , so the first isomorphism theorem yields

$\mathbb Z/ k \mathbb Z = \mathbb Z/{\ker(\phi)} \cong \phi(\mathbb Z) = \langle \zeta_k \rangle.$

Example 2. For any $k > 0$ , $\mathbb R/k \mathbb Z \cong S^1$ .

Proof. Define the surjective homomorphism $\phi : \mathbb R \to \mathbb C^{\times}$ by $\phi(t) = e^{i \cdot 2\pi t/k}$ . Then $\ker(\phi) = k\mathbb Z$ , so the first isomorphism theorem yields

$\mathbb R/ k \mathbb Z = \mathbb R/{\ker(\phi)} \cong \phi(\mathbb R) = S^1.$

Example 3. $\mathbb C^{\times}/S^1 \cong \mathbb R^+$ .

Proof. Define the surjective homomorphism $\phi : \mathbb C^{\times} \to \mathbb R$ by $\phi(z) = |z|$ . Then $\ker(\phi) = S^1$ , so the first isomorphism theorem yields

$\mathbb C^{\times} / S^1 = {\mathbb C^{\times}}/\ker(\phi) \cong \phi(\mathbb C^{\times}) = \mathbb R^+.$

Example 4. $\mathrm{GL}_n(\mathbb F)/\mathrm{SL}_n(\mathbb F) \cong \mathbb F^{\times}$ .

Proof. Define the surjective homomorphism $\phi : \mathrm{GL}_n(\mathbb F) \to \mathbb F$ by $\phi(\mathbf A) = \det(\mathbf A)$ . Then $\ker(\phi) = \mathrm{SL}_n(\mathbb F)$ , so the first isomorphism theorem yields

$\mathrm{GL}_n(\mathbb F) / {\mathrm{SL}_n(\mathbb F)} = { \mathrm{GL}_n(\mathbb F)}/{\ker(\phi)} \cong \phi(\mathrm{GL}_n(\mathbb F)) = \mathbb F^{\times}.$

Lemma 3. Suppose $K \leq G$ and $M \leq G$ . Then $K \cap M \leq K$ . In particular,

$K \cap M \leq MK \leq G.$

Furthermore, if $K \triangleleft G$ , then

$K \triangleleft MK,\quad K \cap M \triangleleft M.$

Proof. Exercise in book-keeping.

Theorem 2. Under the hypotheses of Lemma 3,

$MK/K \cong M/(K \cap M) = (MK \cap M)/(K \cap M).$

This result is known as the second isomorphism theorem, and can be heuristically thought of as “cancelling the $\cap M$ terms”.

Proof. Define the map $\phi : M \to MK/K$ by $\phi(m) = mK$ .

We claim that $\phi$ is a surjective homomorphism with kernel $K \cap M$ .

For the surjection claim, fix $g K \in MK/K$ . Since $g \in MK$ , there exist $k \in K$ and $m \in M$ such that $g = mk$ . Then $m^{-1}(mk) \in K$ implies that

$gK = mkK = mK = \phi(m),$

as required.

For the homomorphism claim, by Lemma 1, the group structure on $(MK)/K$ implies that

$\phi(m_1 m_2) = (m_1 m_2)K = (m_1 K) (m_2 K) = \phi(m_1) \phi(m_2).$

Finally, for the kernel claim, since $\ker(\phi) \subseteq M$ ,

$\begin{aligned} m \in \ker(\phi) \quad &\iff \quad \phi(m) = K \\ &\iff \quad mK = K \\ &\iff \quad m \in K \cap M. \end{aligned}$

Hence, $\ker(\phi) = K \cap M$ . By the first isomorphism theorem,

$M/(K \cap M) = M/{\ker(\phi)} \cong \phi(M) = MK/K.$

Theorem 3. If $K \triangleleft G, M \triangleleft G, K \subseteq M$ , then $M/K \triangleleft G/K$ and

$(G/K)/(M/K) \cong G/M.$

This result is known as the third isomorphism theorem, and can be heuristically thought of as “cancelling the $K$ terms”.

Proof Sketch. Check that the map $\phi : G/K \to G/M$ defined by $\phi(gK) = gM$ is a well-defined surjective homomorphism with kernel $M/K$ , then conclude using the first isomorphism theorem.

Example 5. $(\mathbb Z/6\mathbb Z) / ( 3\mathbb Z/6 \mathbb Z) \cong \mathbb Z/3 \mathbb Z$ .

Definition 2. For $K \leq G$ , define

$\mathrm{Sub}(G, K) := \{ M \in \mathcal P(G) : K \subseteq M \leq G \}$

and $\mathrm{Sub}(G) := \mathrm{Sub}(G, \{e\})$ .

Theorem 4. If $K \triangleleft G$ , then the map $\phi : \mathrm{Sub}(G, K) \to \mathrm{Sub}(G/K)$ defined by $\phi(M) = M/K$ is a well-defined bijection. This result is known as the correspondence theorem.

Proof. We just need to check that $\phi$ is injective and surjective.

For injectivity, suppose $\phi(M_1) = \phi(M_2)$ so that $M_1/K = M_2/K$ . To prove that $M_1 = M_2$ , fix $m \in M_1$ . By definition, $m \in mK \in M_1 K = M_2 K$ . Hence, $m \in M_2$ . The argument works symmetrically.

For surjectivity, fix $H \in \mathrm{Sub}(G/ K)$ . Then there exists $M \subseteq G$ such that

$H = \{mK : m \in M\}.$

It suffices to check that $K \subseteq M \leq G$ . Since $H \leq G/K$ , it contains the identity element $K$ . In particular, $K = kK \in H$ , so that $k \in M$ , yielding $K \subseteq M$ . For the subgroup property, fix $a, b \in M$ . Since $K \triangleleft G$ and $H \leq G/K$ ,

$(ab^{-1})K = (aK)(b^{-1}K) = (aK)(bK)^{-1} \in H.$

Therefore, $ab^{-1} \in M$ . Hence, $M \leq G$ , as required. Finally, by construction,

$\phi(M) = M/K = H,$

as required.

These ideas in group isomorphisms are general, but become even more powerful when used in tandem with the more finite aspects of group theory. We will need to explore these ideas next time, starting with the symmetric groups—the set of bijections on a set.

Eventually, we will discuss modulo arithmetic using this group-theoretic lens, yielding natural theorems that would otherwise be nontrivial without group-theoretic language. In fact, we have encountered the desired set(s); it is simply $\mathbb Z_k = \mathbb Z/k\mathbb Z$ , but together with an associated multiplication operation.

—Joel Kindiak, 29 Dec 25, 1416H

April 23, 2026