CHAPTER 08 · MA-0561

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Normal Subgroups

Normal subgroups

Definition (Product of sets)

Let \(G\) be a group and \(S,T \subseteq G\). Define \(ST = \{ s \ast t: s \in S, t \in T \}\). Note that if \(S = \{ s \}\) and \(T\leq G\), then \(sT =ST\), a coset of \(G\). Note that if \(S,T,U \subseteq G\), . If \(\mathcal{F} = \{ A:A\subseteq G \}\), then \(\cdot\) is defined on \(\mathcal{F}\) and is associative.

Definition (Normal subgroup)

Let \(H \leq G\). We say that \(H\) is a normal subgroup of \(G\) and we write \(H \triangleleft G\) if \(Hx = xH\) for every \(x \in G\).

Note (Properties of normal subgroups)

  1. If \(G\) is abelian, every subgroup is normal.
  2. \(xH = Hx \iff xH x^{-1} = \{ x \ast h \ast x ^{-1}: h \in H \} = H\).
  3. If \(H \triangleleft G\), this does not imply that for all \(x \in G\) and \(h \in H\) \(x \ast h = h \ast x\). What is true is that if \(h \in H\), there exists \(h' \in H\) such that \(x \ast h = h' \ast x\).

  4. If for every \(x \in G\), \(x H x ^{-1} \subseteq H\), this necessarily implies equality of the two sets, and hence that \(H \triangleleft G\). Suppose that \(xHx ^{-1} \subseteq H\) for every \(x \in G\). Note that, for \(x \in G\).

    \[H = x ^{-1}(\underbrace{ x H x ^{-1} }_{ \in H }) x \subseteq x ^{-1} H x \subseteq H \implies x ^{-1} H x = H.\]
  5. For every group \(G\), \(\{ 1_{G} \} \triangleleft G\), \(G \triangleleft G\).

Proposition (Chained subgroups)

Let \(K,H,G\) be groups with \(K\leq H\leq G\). If \(K \triangleleft G\), then \(K \triangleleft H\).

Proof: The result is trivial, since \(G \subseteq H\).

Theorem (The kernel is a normal subgroup)

Let \(f:G\to G'\) be a group homomorphism. Then \(\operatorname{Ker}(f) \triangleleft G\).

Proof: Let \(H:= \operatorname{Ker}(f)\). Let \(x \in G\). We must show that \(xHx ^{-1} \subseteq H\). Let \(h \in H\). Then \(f(x \ast h \ast x ^{-1}) = f(x) \ast f(h) \ast f(x ^{-1}) = 1_{G'}\).

Example (Normal subgroup in \(S_3\))

Let \(G = S_{3} = \{ (1), \begin{pmatrix}1 & 2\end{pmatrix}, \begin{pmatrix}1 & 3\end{pmatrix}, \begin{pmatrix}2 & 3\end{pmatrix} \begin{pmatrix}1 & 2 & 3\end{pmatrix} \begin{pmatrix}3 & 2 & 1\end{pmatrix} \}\). Consider \(H = \langle \begin{pmatrix}1 & 2 & 3\end{pmatrix} \rangle\). We have that

and similarly for the remaining cycles.