Section8.2Focusing on normal subgroups

We first provide a theorem that will help us in identifying when a subgroup of a group is normal. First, we provide a definition.

Definition8.2.1

Let $H$ be a subgroup of $G$ and $a,b$ in $G\text{.}$ We define

\begin{equation*} aHb=\{ahb\,:h\in H\}. \end{equation*}
Proof

We now consider some examples and nonexamples of normal subgroups of groups.

Example8.2.3

1. As previously mentioned, the trivial and improper subgroups of any group $G$ are normal in $G\text{.}$

2. As previously mentioned, if group $G$ is abelian then each of its subgroups is normal in $G\text{.}$

3. Suppose $H\leq G$ has $(G:H)=2\text{.}$ Then $H \unlhd G\text{.}$ The proof of this is left as an exercise for the reader.

4. Example 7.2.9 shows that subgroup $H=\langle (12)\rangle$ isn't normal in $S_3$ (for example, $(13)H\neq H(13)\text{.}$ But $\langle (123)\rangle$ must be normal in $S_3$ since $(S_3:\langle (123)\rangle )=6/3=2.$

5. $\langle r\rangle$ is normal in $D_n$ since $(D_n:\langle r\rangle )=2\text{.}$

6. $\langle f\rangle$ isn't normal in $D_4\text{:}$ for instance,

\begin{equation*} r\langle f\rangle r^{-1}=\{e,rfr^3\}=\{e, fr^3r^3\}=\{e,fr^2\}\not\subseteq \langle f\rangle . \end{equation*}

We consider two other very significant examples. First, we revisit the idea of the center of a group, first introduced in Example 2.8.9.

Definition8.2.4

Let $G$ be a group. We let

\begin{equation*} Z(G)=\{z\in G\,:\, az=za \text{ for all } a\in G\}. \end{equation*}

$Z(G)$ is called the center of $G\text{.}$

(The Z stands for “zentrum,” the German word for “center.”)

Proof

The next definition is profoundly important for us.

Definition8.2.7

Let $G$ and $G'$ be groups and let $\phi$ a homomorphism from $G$ to $G'\text{.}$ Letting $e'$ be the identity element of $G'\text{,}$ we define the kernel of $\phi$, $\Ker \phi\text{,}$ by

\begin{equation*} \Ker \phi = \{k\in G\,:\,\phi(k)=e'\}. \end{equation*}
Proof

One slick way, therefore, of showing that a particular set is a normal subgroup of a group $G$ is by showing it's the kernel of a homomorphism from $G$ to another group.

Example8.2.9

Let $n\in \Z^+\text{.}$ Here is a rather elegant proof of the fact that $SL(n,\R)$ is a normal subgroup of $GL(n,\R)\text{:}$ Define $\phi: GL(n,\R) \to \R^*$ by $\phi(A)=\det A\text{.}$ Clearly, $\phi$ is a homomorphism, and since the identity element of $\R^*$ is 1,

\begin{equation*} \Ker \phi=\{A\in GL(n,\R)\,:\,\det A= 1\}=SL(n,\R). \end{equation*}

So $SL(n,\R)\unlhd GL(n,\R)\text{.}$