The Quotient Group

Proposition (Normal groups and operations)

If \(H \triangleleft G\) and \(a,b \in G\), then \(aH \cdot bH = (a \ast b) H\)

Proof: Note that

\[\begin{aligned} aH \cdot bH \end{aligned} = aH \cdot Hb = a Hb = a b H = (a \ast b) H.\]

Theorem (Existence of the quotient group)

Let \(H \triangleleft G\). Then \(G / H\) is a group under the operation \(\cdot\)

\[a H \cdot bH = (a \ast b) H.\]

Proof: The operation is well defined and associative (see the previous proposition).

Membership of the identity: \(1_{G} \cdot H = H\)
Existence of inverses: \(aH \cdot a^{-1}H = (a \ast a^{-1}) H = H = id_{G / H }\).

Note (Independence of the representative)

Why does it not depend on the representative? Suppose that \(aH = \hat{a}H\) and \(bH = \hat{b}H\). Then

\[aH \cdot bH = \hat{a} H \cdot \hat{b}H \iff (a \ast b) H = (\hat{a} \ast \hat{b}) H \iff (a \ast b) \ast (\hat{ a} \ast \hat{b})^{-1} \in H,\]

which is clearly true.

Theorem (Every normal subgroup is the kernel of a homomorphism)

Let \(H \triangleleft G\) and consider \(f:G\to G / H\) such that \(f(a) = a H\). Then \(f\) is an epimorphism and \(\operatorname{Ker}(f) = H\).

Proof: We shall first show that \(f\) is a homomorphism. Note that \(f(a \ast b) = (a \ast b) H = aH \cdot bH = f(a) \cdot f(b)\). Surjectivity follows from the definition. Now, if \(x \in H\), then \(f(x) = xH = H =id_{G / H}\), hence, \(x \in \operatorname{Ker} f\). If \(x \in \operatorname{Ker} f\), then \(f(x) = xH = H\), hence \(x \in H\).

Example (Trivial quotients)

Note that \(\{ 1_{G} \} \triangleleft G \implies G / \{ 1_{G} \} = \{ a \cdot \{ 1_{G}: a \in G \} \} = \{ \{ a \}: a \in G \} \cong G\).
\(G \triangleleft G \implies G / G = \{ a \cdot G: a \in G \} = \{ G \} = \{ 1_{G / G} \}\).

Example (Quotients in \(\mathbb{Z}\) and in a cyclic group)

Let \(G = (\mathbb{Z}, + )\), \(H = 4\mathbb{Z}\). Note that \(H \triangleleft G\) since \(G\) is abelian. Note that \(G / H \cong \mathbb{Z}_{4}\), since \(aH = bH \iff a + (-b) \in 4\mathbb{Z} \iff 4 \mid a-b \iff a \cong b \quad \text{mod }4\).
Let \(G\) be a group and let \(a \in G\) be such that \(\lvert a \rvert = 8\). and \(G=\langle a \rangle\). Let \(H = \langle a^{4}. \rangle\) Since \(G\) is commutative. \(H \triangleleft G\). We have that \(H = \{ 1_{G, }a^{4} \}\) and \(G = \{ 1_{G}, a, a^{2},\dots,a^{7} \}\). What is \(G/H\)? Let us find the cosets:
1. \[H = a^{4}H\]
2. \[aH = \{ a, a^{5} \} = a^{5} H\]
3. \[a^{2}H = \{ a^{2}, a^{6} \} = a^{6}H\]
4. \(a^{3}H = \{ a^{3},a^{7} \}= a^{7}H\) Thus, \(G/H = \{ H, aH, a^{2}H, a^{3}H \}\) and \(G = H \hspace{2mm} \dot{\cup} \hspace{2mm} aH \hspace{2mm} \dot{\cup} \hspace{2mm} a^{2}H \hspace{2mm}\dot{\cup} \hspace{2mm} a^{3}H\).

Definition (Commutator and commutator subgroup)

Let \((G,\ast)\) be a group. Let \(a,b \in G\). Define the commutator of \(a,b\) as

\[[a,b] = a \ast b \ast a^{-1} \ast b ^{-1} \in G\]

Define the commutator subgroup of \(G\) as \(G':= \langle [a,b]:a,b \in G \rangle\).

Exercise (Characterisation via conjugation)

Let \(S < G\). Then \(S \triangleleft G\) if and only if \(C_{a}(S) \leq S\) for every \(a \in G\).

Theorem (Commutator subgroup and abelian quotients)

Let \(G\) be a group and \(G'\) its commutator subgroup. Then \(G' \triangleleft G\). Moreover, if \(H \triangleleft G\), then \(G/H\) is abelian if and only if \(G' \triangleleft H\).

Proof: We must show that, for every \(g \in G\), \(gG'g^{-1} \subseteq G'\). It suffices to show that if \(a, b \in G\), \(g \in G\) , then \(g[a,b]g^{-1} \in G'\), why? If \(z = [a_{1} , b_{1}] [a_{2}, b_{2}]\), then

\[gzg^{-1} = g [a_{1},b_{1}] [a_{2}, b_{2}] g^{-1} = (g[a_{1},b_{1}]g^{-1})(g [a_{2},b_{2}] g^{-1}),\]

hence, it suffices to prove the result for an arbitrary commutator and it follows immediately for every \(z \in G'\), Let \([a,b] \in G'\). Then,

\[\begin{aligned} g \ast a \ast b \ast a^{-1} \ast b^{-1} \ast g^{-1} \end{aligned}\] \[= (g \ast a \ast g^{-1}) \ast (g \ast b \ast g^{-1}) \ast(g \ast a^{-1} \ast g ^{-1}) \ast (g \ast b ^{-1} \ast g ^{-1}) = [g \ast a \ast g ^{-1},g \ast b \ast g ^{-1} ]\in G'.\]

Suppose now that \(G/H\) is abelian. Then for all \(a,b \in G\), \(aH \cdot b H = b H \cdot a H\), so that \((a \ast b) H = (b \ast a)H\). Hence , \((a \ast b) \ast (b \ast a) ^{-1} = a \ast b \ast a^{-1} \ast b ^{-1} = [a,b] \in H\). Thus, \(\{ [a,b]: a,b \in G \} \subseteq H \implies G' \subseteq H \implies G' \leq H\). (Write out the other direction).

Example (\(A_3 \triangleleft S_3\))

Consider \(S_{3} = \{ (1),\begin{pmatrix}1 & 2\end{pmatrix}, \begin{pmatrix}2 & 3\end{pmatrix}, \begin{pmatrix}1 & 3\end{pmatrix}, \begin{pmatrix}1 & 2 & 3\end{pmatrix} \begin{pmatrix}3 & 2 & 1\end{pmatrix} \}\) and \(A_{3} = \{ \sigma \in S_{3}: \sigma \text{ is even} \} = \{ \begin{pmatrix}1\end{pmatrix}, \begin{pmatrix}1 & 2 & 3\end{pmatrix}, \begin{pmatrix}3 & 2 & 1\end{pmatrix}\}\). We shall show that \(A_{3} \triangleleft S_{3}\). Consider \(S = \{ -1,1 \}\). Note that \((S, \cdot) = (\mathbb{Z}_{2}, +)\). Let \(f:S_{3} \to S\) be such that \(\sigma \mapsto \operatorname{sgn}( \sigma )\). Note that \(f\) is a homomorphism, since

\[\begin{aligned} f(\sigma_{1} \circ \sigma_{2}) &= \operatorname{sgn}(\sigma_{1} \circ \sigma_{2} )\\ &= \operatorname{sgn}( \sigma_{1} ) \cdot \operatorname{sgn}( \sigma_{2} )\\ &= f(\sigma_{1}) \cdot f(\sigma_{2}). \end{aligned}\]

Note that \(\operatorname{Ker} f = A_{3}\), from which we conclude that \(A_{3} \triangleleft S_{3}\). What is \(S_{3}/A_{3}\)? Let us construct the cosets

\[A_{3}\]
\(\begin{pmatrix}1 & 2\end{pmatrix} A_{3} = \{ \begin{pmatrix}1 & 2\end{pmatrix}, \begin{pmatrix}3 & 2\end{pmatrix}, \begin{pmatrix}1 & 3\end{pmatrix} \}\).

Thus, \(S_{3}/A_{3} = \{ A_{3}, \begin{pmatrix}1&2\end{pmatrix}A_{3} \} \cong \mathbb{Z}_{2}\).

Example (\(\langle r_1\rangle \triangleleft D_4\))

Consider \(D_{4} = \{ r_{0},r_{1},r_{2},r_{3},s_{1},s_{2},s_{3},s_{4} \}\) the group of symmetries of the square and \(H = \langle r_{1} \rangle = \{ r_{0},r_{1},r_{2},r_{3} \}\). We must show that for every \(i \in \{ 1,\dots,4 \}\), \(s_{i} H s_{i} ^{-1} \subseteq H\). Note first that for every \(i\), \(s_{i}^{2} = r_{0}\), hence, \(s_{i} = s_{i}^{-1}\). Hence, we must show that for all \(j \in \{ 0,1,2,3 \}, i \in\{ 1,2,3,4 \}\), \(s_{i} r_{j} s_{i} ^{-1} \in H\)