Motivation

As I often forget how to prove the second isomorphism theorem in group theory, here is one (in all painstaking details) for the record.

Theorem statement

Given group \(G\), subgroup \(S\leq G\), normal subgroup \(N\trianglelefteq G\), then

(1) \(SN \leq G\)
(2) \(S\cap N \trianglelefteq S\)
(3) \((SN)/N \cong S/(S\cap N)\)

Proof of (1)

First, to show that \(SN\) is a subset. \(SN=\{sn | s\in S, n\in N\}\). Since \(s\in G, n \in G\), by closure of group multiplication, \(sn\in G\), so \(SN\) is a subset of \(G\).

Next, we need to show that \(SN\) is itself a group. Check the group axioms:

Identity: \(e\) is the identity element in \(SN\).
Inverse: Given \(sn \in SN\), then its inverse is \(n^{-1}s^{-1}\). Let \(n_3 = sn^{-1}s^{-1}\); since \(N\) is normal, \(n_3\in N\). Then the inverse element \(n^{-1}s^{-1}= s^{-1}n_3\), which is in \(SN\).
Associativity: Since \(s_in_i\) is an element in \(G\), by associativity of \(G\), \((s_1n_1\cdot s_2n_2)\cdot s_3n_3 = s_1n_1\cdot (s_2n_2\cdot s_3n_3)\)
Closure: Need to show \(s_1n_1s_2n_2 \in SN\). Let \(n_3 = s_2^{-1}n_1s_2\); by normality of \(N\), \(n_3\in N\). Then \(s_1n_1s_2n_2 = s_1 s_2 n_3 n_2\), which is manifestly in \(SN\).

Proof of (2)

First, \(S\cap N\) is obviously a subset of \(S\).

Next, to show that \(S\cap N\) is a group. Again check the group axioms:

Identity: \(e\in S\) and \(e\in N\), so identity exists.
Inverse: given \(k\in S\cap N\), \(k\in S\) implies \(k^{-1} \in S\); \(k \in N\) implies \(k^{-1} \in N\); hence \(k^{-1}\in S\cap N\).
Associativity: Obvious as group elements of \(G\).
Closure: Given \(k,p \in S\cap N\), then \(k\in S\) and \(p \in S\) implies \(kp \in S\); \(k\in S\) and \(p \in S\) implies \(kp \in N\). So \(kp \in S\cap N\).

Last, to show that \(S\cap N\) is normal with respect to \(S\), need to show \(\forall k \in S\cap N, \forall s\in S, sks^{-1} \in S\cap N\). Since both \(s,k\in S, sks^{-1}\in S\). Since \(k\in N, s\in G\) and \(N \trianglelefteq G\), then \(sks^{-1} \in N\). So \(sks^{-1} \in S\cap N\).

Proof of (3)

In order to form the quotient group \((SN)/N\), need to first show that \(N \trianglelefteq SN\). We already know \(N\) is a subset of \(SN\), and it is a group, so it remains to show the normality condition. \(\forall n'\in N, sn \in SN\), need to show \((sn) n' (sn)^{-1}=snn'n^{-1}s^{-1} \in N\). First insert pairs of \(s^{-1}s\) in between every elements of \(N\) to obtain \((s n s^{-1}) (s n' s^{-1})(s n^{-1} s^{-1})\). Since \(N \trianglelefteq G\), all three parathesized elements are in \(N\), and then use closure to conclude \((sn) n' (sn)^{-1}\in N\).

Next, need to understand the group element in \((SN)/N\) and \(S/(S\cap N)\). \((SN)/N=\{[[sn] \in SN\}\), where the equivalence relation is given by \(s_1n_1 \sim s_2 n_2\) if \(\exists n\) such that \(s_1n_1 = n \cdot s_2n_2\).

One strategy is to construct a homomorphism \((SN)/N \rightarrow S/(S\cap N)\), then show its inverse exists. Or show it is both injective and surjective. This requires quite a lot of low-level detail defining the homormophisms and proving their properties.

A better strategy is to use the first isomorphism theorem: given a homomorphism \(h: G\rightarrow H\), then \(G/\text{ker}(h) = \text{Im}(h)\). I can choose a surjective homormophism such that either of

\(h: SN \rightarrow S/(S\cap N)\) such that \(\text{ker}(h)=N, \text{Im}(h) = S/(S\cap N)\)
\(h: S \rightarrow (SN)/N\) such that \(\text{ker}(h)=S\cap N, \text{Im}(h) = (SN)/N\)

We will take the first option. Define the homomorphism \(h: SN \rightarrow S/(S\cap N)\) by \(h(sn) = [s]\), with the equivalence relation \(s_1\sim s_2\) if \(\exists n\in N\) such that \(s_1=n s_2\). This homomorphism is surjective since \(\forall [s]\in S/(S\cap N), \exists s\in S\) such that \(h(s)=[s]\). To find the kernel of the homomorphism, need to know what is in the equivalence class \([e]\), which by definition of \(S/(S\cap N)\), is \(N\). Then \(\forall n\in N, h(n) = [e]\). For \(s \in SN, s\notin N, h(s)=[s]\neq [e]\). Therefore, \(\text{ker}(h)=N\).

Q.E.D.