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*}We first provide a theorem that will help us in identifying when a subgroup of a group is normal. First, we provide a definition.
Let \(H\) be a subgroup of \(G\) and \(a,b\) in \(G\text{.}\) We define
\begin{equation*} aHb=\{ahb\,:h\in H\}. \end{equation*}Let \(H\) be a subgroup of a group \(G\text{.}\) Then the following are equivalent:
\(H\) is normal in \(G\text{;}\)
\(aHa^{-1}=H\) for all \(a\in G\text{;}\)
\(aHa^{-1}\subseteq H\) for all \(a\in G\text{.}\)
We now consider some examples and nonexamples of normal subgroups of groups.
As previously mentioned, the trivial and improper subgroups of any group \(G\) are normal in \(G\text{.}\)
As previously mentioned, if group \(G\) is abelian then each of its subgroups is normal in \(G\text{.}\)
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.
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 \((S_3:\langle (123)\rangle )=6/3=2.\)
\(\langle r\rangle\) is normal in \(D_n\) since \((D_n:\langle r\rangle )=2\text{.}\)
\(\langle f\rangle\) 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.
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.”)
Let \(G\) be a group. Then \(Z(G)\) is a subgroup of \(G\text{.}\)
Let \(G\) be a group and let \(H\) be a subgroup of \(G\) with \(H\subseteq Z(G)\text{.}\) Then \(H\unlhd G\text{.}\) In particular, \(Z(G)\) is itself a normal subgroup of \(G\text{.}\)
The next definition is profoundly important for us.
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*}Let \(G\) and \(G'\) be groups and let \(\phi\) be a homomorphism from \(G\) to \(G'\text{.}\) Then \(\Ker \phi\) is a normal subgroup of \(G\text{.}\)
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.
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 \(SL(n,\R)\unlhd GL(n,\R)\text{.}\)