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

KindiakMath

Group Isomorphisms

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Group Isomorphisms

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