Finite Richness

Our discussions on group theory thus far have made no restriction on whether the groups in discussion have finitely many elements or not. By insisting that these groups are finite, we end up with even more fascinating results.

Let’s first talk about roots of polynomials. In a sense, that’s where our discussions really began. Given any positive integer $k$ , the fundamental theorem of algebra asserts that the equation

$z^k = 1,\quad z \in \mathbb C,$

has exactly $k$ complex (possibly repeated) roots. Furthermore, regardless of $k$ , the intermediate value theorem asserts that the equation $z^k = 1$ has at most $2$ real roots. Hence, for any $k \geq 3$ , there must exist at least one complex non-real solution to the equation $z^k =1$ .

Theorem 1. Fix any positive integer $k \geq 1$ . Define the primitive $k$ -th root by $\zeta_k := e^{2\pi i /k}$ . Then the set of solutions to the equation $z^k = 1$ is given by

$\langle \zeta_k \rangle := \{ 1, \zeta_k, \zeta_k^2, \dots, \zeta_k^{k-1}\}.$

Furthermore, $\langle \zeta_k \rangle \leq S^1$ under complex number multiplication.

Proof. It is clear that $\zeta_k^k = e^{2\pi i} = 1$ and that for each $\ell = 0, 1, \dots, k-1$ , $\zeta_k^{\ell} \neq 1$ .

We first check that $\langle \zeta_k \rangle$ has $k$ elements. Given $0 \leq \ell, m < k$ ,

$\displaystyle \zeta_k^{\ell-m} = g(\ell-m) = \frac{g(\ell)}{g(m)} = 1$

implies that $\ell - m = 0 \Rightarrow \ell = m$ . Taking the contrapositive, for $\ell \neq m$ , $\zeta_k^{\ell} \neq \zeta_k^m$ . Therefore, the map

$g : \{0,1,\dots,k-1\} \to \langle \zeta_k \rangle,\quad g(\ell) = \zeta_k^{\ell}$

is bijective, so that $\langle \zeta_k \rangle$ has $k$ elements.

We next check that each $\zeta_k^{\ell} \in \langle \zeta_k \rangle$ is a solution to the equation $z^k = 1$ :

$(\zeta_k^{\ell})^k = (\zeta_k^k)^{\ell} = 1^{\ell} = 1.$

By the fundamental theorem of algebra, since $z^k = 1$ has exactly $k$ solutions, $\langle \zeta_k \rangle$ gives all $k$ solutions to the equation $z^k = 1$ .

Finally, define the group homomorphism $f : \mathbb Z \to S^1$ by $f(n) = \zeta_k^n$ . Then $\langle \zeta_k \rangle \subseteq f(\mathbb Z)$ since $\zeta_k^\ell = f(\ell)$ . For any integer $n$ , find unique integers $q, r$ such that

$n = qk + r,\quad 0 \leq r < k.$

Recalling that $f(k) = \zeta_k^k = 1$ ,

$\begin{aligned} f(n) &= f(qk+r) \\ &= (f(k))^q \cdot f(r) \\ &= 1^q \cdot \zeta_k^r \\ &= \zeta_k^r \in \langle \zeta_k \rangle. \end{aligned}$

Hence, $f(\mathbb Z) \subseteq \langle \zeta_k \rangle$ . In particular, $\langle \zeta_k \rangle = f(\mathbb Z) \leq S^1$ .

The group $\langle \zeta_k \rangle$ is called the $k$ -th roots of unity, and has a very rich group-theoretic structure applicable across mathematical inquiry.

Definition 1. Let $G$ be a group. For any $x \in G$ , define the cyclic group generated by $x$ by

$\langle x \rangle := \{ x^n : n \in \mathbb Z \}.$

It is clear that $\langle x \rangle \leq G$ . We say that $G$ is cyclic if there exists $y \in G$ such that $G = \langle y \rangle$ .

Example 1. Defining the homomorphism $f : \mathbb Z \to S^1$ by $f(n) = \zeta_k^n$ , since

$\begin{aligned} \langle \zeta_k \rangle &= \{ \zeta_k^n : n \in \mathbb Z\}\\ &= \{f(n) : n \in \mathbb Z\} \\ &= f(\mathbb Z) = \{ 1, \zeta_k, \zeta_k^2, \dots, \zeta_k^{k-1}\}. \end{aligned}$

the group $\{ 1, \zeta_k, \zeta_k^2, \dots, \zeta_k^{k-1}\}$ is indeed cyclic, and the notation

$\langle \zeta_k \rangle = \{ 1, \zeta_k, \zeta_k^2, \dots, \zeta_k^{k-1}\}$

is justified.

Example 2. Under addition, $\mathbb Z = \langle 1 \rangle$ is cyclic. However, $\mathbb Q$ is not cyclic.

Proof. Suppose for a contradiction that $\mathbb Q$ is cyclic. Write $\mathbb Q = \langle r \rangle$ for some $r > 0$ without loss of generality. Then $r/2 \in \mathbb Q \backslash \langle r \rangle$ since $r/2 \in \langle r \rangle$ implies

$\displaystyle \frac r2 = kr \quad \Rightarrow \quad \frac 12 = k \in \mathbb Z,$

a blatant contradiction.

Are subgroups of cyclic groups themselves cyclic? Yes!

Theorem 1. Let $G$ be a group and $H \leq G$ . If $G$ is cyclic, then so is $H$ .

Proof. Since $G$ is cyclic, there exists $x \in G$ such that $G = \langle x \rangle$ . Since $G$ is cyclic, there exists an integer $n$ such that $y = x^n$ . Since $H \leq G$ , $x^{-n} = y^{-1} \in H$ . Hence, without loss of generality, let $n$ be the smallest positive integer such that $y = x^n$ .

It is clear that $\langle y \rangle \subseteq H$ . To show that $H \subseteq \langle y \rangle$ , we first fix $z \in H \subseteq \langle x \rangle$ . Without loss of generality, there exists $\ell \in \mathbb Z^+$ such that $z = x^{\ell}$ . By the division algorithm, there exists unique integers $q , r$ such that

$\ell = qn + r,\quad 0 \leq r < n.$

In particular,

$z = x^{\ell} = x^{qn} x^r = (x^n)^q x^r = y^q x^r.$

Since $H \leq G$ , $x^r = z (y^{q})^{-1} \in H$ . Since $r < n$ and $n$ is the smallest positive integer such that $x^n \in H$ , we must have $r = 0$ , so that $x^r = x^0 = e$ . Hence,

$z = y^q \in \langle y \rangle.$

Remark 1. When there is no ambiguity, we abbreviate $(G, *) \equiv G$ and $x * y = xy$ .

Cyclic groups are quite fascinating objects. It turns out that all cyclic groups are, from the lens of group theory, not much different from either $\langle \zeta_k \rangle$ or $\mathbb Z$ . We can define “not much different” using the notion of group isomorphisms.

Lemma 1. Let $f : G \to H$ be a group homomorphism. Then $f$ is injective if and only if $\ker(f) = \{e\}$ .

Proof. $(\Rightarrow)$ Since $f$ is a homomorphism, $\{e\} \subseteq \ker(f)$ . For any $x \in \ker(f)$ ,

$f(x) = e = f(e).$

Since $f$ is injective, $x = e$ . Hence, $\ker(f) \subseteq \{e\}$ .

$(\Leftarrow)$ Fix $x, y \in G$ . Suppose $f(x) = f(y)$ . Since $f$ is a homomorphism,

$f(xy^{-1}) = f(x) f(y)^{-1} = e.$

Hence, $xy^{-1} \in \ker(f) = \{e\}$ implies that

$xy^{-1} = e \quad \Rightarrow \quad x = y.$

Hence, $f$ is injective.

Definition 2. A group homomorphism $f : G \to H$ is said to be a group isomorphism if it is bijective. In this case, we write $G \cong H$ , and remark the following equivalence relation properties, left as an exercise:

For any group $G$ , $G \cong G$ .
For groups $G, H$ , $G \cong H$ implies $H \cong G$ .
For groups $G, H, K$ , $G \cong H$ and $K \cong K$ implies $G \cong K$ .

Example 3. For any $a > 0$ , define $\langle a \rangle \leq \mathbb R^{+}$ . Then $\langle a \rangle \cong \mathbb Z$ via the isomorphism $f : \mathbb Z \to \langle a \rangle$ defined by $f(n) = a^n$ .

Theorem 2. Let $G$ be a group. If $G$ is cyclic, then either $G \cong \langle \zeta_k \rangle$ or $G \cong \mathbb Z$ . The converse holds trivially.

Proof. Fix $x \in G$ such that $G = \langle x \rangle$ . There are two cases to check:

There exists $k \in \mathbb N^+$ such that $x^k = e$ .
For any $n \in \mathbb N^+$ , $x^n \neq e$ .

In the first case, suppose $k$ is minimum without loss of generality. Define the well-defined surjective homomorphism $f : \langle x \rangle \to \langle \zeta_k \rangle$ by $f(x^\ell) = \zeta_k^\ell$ . We note that this definition is well-defined since for any $\ell = qn + r$ with $0 \leq r < n$ ,

$f(x^\ell) = f(x^n)^q f(x^r) = f(e)^q f(x^r) = f(x^r).$

Furthermore, $f(x^r) = 1$ implies $r = 0$ , since $n$ is minimal, so that $x^r = x^0 = e$ , which implies that $f$ is injective. Hence, $f$ is a group isomorphism, and $G = \langle x \rangle \cong \langle \zeta_k \rangle$ .

In the second case, define the surjective homomorphism $f: \mathbb Z \to \langle x \rangle$ by $f(n) = x^n$ . To check injectivity, we note that

$f(n) = 1 \quad \Rightarrow \quad x^n = 1 \quad \Rightarrow \quad n = 0.$

Hence, $\ker(f) = \{0\}$ , so that $f$ is injective by Lemma 1. Hence, $f$ is a group isomorphism, and $\langle x \rangle \cong \mathbb Z$ .

Lemma 2. Let $f : G \to H$ be a surjective group homomorphism. Then $H$ is Abelian if $G$ is.

Proof. S Fix $x, y \in H$ . Since $f$ is surjective, there exists $x_0, y_0 \in G$ such that $x = f(x_0)$ and $y = f(y_0)$ . Since $G$ is Abelian,

$xy = f(x_0) f(y_0) = f(x_0 y_0) = f(y_0 x_0) = f(y_0) f(x_0) = yx.$

Hence, $H$ is Abelian.

Corollary 1. All cyclic groups are Abelian.

Proof. Let $G$ be any cyclic group. By Theorem 2, $G \cong \langle \zeta_k \rangle$ or $G \cong \mathbb Z$ , and both groups are Abelian. In the former, there exists a bijection $f : \langle \zeta_k \rangle \to G$ , which is surjective, so that $G$ is Abelian by Lemma 2. The latter case follows similarly by replacing $\langle \zeta_k \rangle$ with $\mathbb Z$ .

Now in the case $G \cong \mathbb Z$ , $|G|$ is infinite. In the case $G = \langle x \rangle \cong \langle \zeta_k \rangle$ , $|G| = k$ , in which case we call $G$ a finite group.

Lemma 3. Let $G$ be a group and $H \leq G$ . For any $x \in G$ , define the left-coset by

$xH := \{x y : y \in G\}.$

Then $\{xH : x \in G\}$ forms a partition of $G$ in the following sense:

For any $x \in G$ , $x H \neq \emptyset$ .
For any $x_1, x_2 \in G$ , $x_1 H \cap x_2 H \neq \emptyset \iff x_1 H = x_2 H$ .
$G = \bigcup_{x \in G} xH$ .

Furthermore, for any $x \in G$ , there exists a bijection $f : H \to x H$ .

Proof. Define the equivalence relation $\sim$ on $G$ by

$x \sim y \quad \iff \quad y^{-1} x \in H \iff \quad x \in yH.$

Then the equivalence relation induces a partition $G/{\sim} =: G/H$ on $G$ . Furthermore, the elements $[x]$ of $G/H$ are described by

$\begin{aligned} [x] &= \{z \in G : z \sim x\} = \{z \in G : z \in xH\} = xH. \end{aligned}$

For the bijection claim, define $f : H \to xH$ by $f(y) = xy$ . It is clear that $f$ is surjective, and for injectivity,

$f(y_1) = f(y_2) \quad \Rightarrow \quad xy_1 = xy_2 \quad \Rightarrow \quad y_1 = y_2.$

Corollary 2. Define the integers modulo $k$ by $\mathbb Z_k := \mathbb Z/k\mathbb Z$ , corresponding to the equivalence relation

$m \sim n \quad \iff \quad m -n \in k\mathbb Z.$

Then

$\mathbb Z_k = \{ [0], [1], [2], \dots, [k-1] \},\quad [m] = m+k\mathbb Z.$

Furthermore, the map $f : \mathbb Z_k \to \langle \zeta_k \rangle$ defined by $[m] \mapsto \zeta_k^m$ is a well-defined bijection, which motivates the definition of addition in $\mathbb Z_k$ by

$[m] + [n] := f^{-1}(f([m]) \cdot f([n])) = [m+n].$

Then $\mathbb Z_k$ can be shown to form a group under addition in $\mathbb Z_k$ , turning $f$ into a group isomorphism, and $\mathbb Z_k \cong \langle \zeta_k \rangle$ . By Theorem 2, $\mathbb Z_k$ is cyclic, and thus Abelian.

Theorem 3 (Lagrange). Let $G$ be a finite group. For any $H \leq G$ , $H$ is a finite group, and $G/H$ is a finite set. Define $[G : H] := |G/H|$ . Then $|G| = [G:H] \cdot |H|$ .

Proof. If $G$ is finite, then the union of cosets must be a finite union. Due to the partition property in Lemma 3, suppose without loss of generality that $n = |G/H| = [G:H]$ so that

$\{H, x_1 H, \dots, x_{n-1} H\}$

partitions $G$ :

$\displaystyle G = \bigsqcup_{i=0}^{n-1} x_i H,$

where we defined $x_0 = e$ for convenience. Since for each $i$ , there exists a bijection $f : H \to x_i H$ as per Lemma 3,

$\displaystyle |G| = \sum_{i=1}^n |x_i H| = \sum_{i=1}^n |H| = n |H| = [G : H] \cdot |H|.$

In particular, $|H| \mid |G|$ .

Corollary 3. Let $G$ be a finite group. Then for any $x \in G$ , there exists a smallest positive $n \mid |G|$ such that $x^n = e$ . In fact, $n = |\langle x \rangle|$ . In particular, $x^{|G|} = e$ .

$\langle x \rangle \cong \langle \zeta_n \rangle.$

Hence, the smallest positive $\ell$ such that $x^\ell = e$ is $\ell = n$ . Writing $m := |G|/n \in \mathbb N^+$ ,

$x^{|G|} = x^{m n} = (x^n)^{m} = e^{m} = e.$

Referring to Corollary 2, recall the usual surjective homomorphism $\phi : \mathbb Z \to \langle \zeta_k \rangle$ by $\phi(n) = \zeta_k^n$ . By checking that $\ker(\phi) = k\mathbb Z$ , we can restate the isomorphism in Corollary 2 as