8.2: Focusing 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.
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*}
Theorem \(\PageIndex{1}\)
Let \(H\) be a subgroup of a group \(G\text{.}\) Then the following are equivalent:
-
\(H\) is normal in \(G\text{;}\)
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\(aHa^{-1}=H\) for all \(a\in G\text{;}\)
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\(aHa^{-1}\subseteq H\) for all \(a\in G\text{.}\)
- Proof
-
The equivalence of Statements 1 and 2 is clear, as is the fact that Statement 2 implies Statement 3. So it suffices to show that Statement 3 implies Statement 2. Well, assume that \(aHa^{−1}⊆H\) for all \(a∈G\), and let \(b∈G\). We want to show that \(bHb^{−1}=H\). Since Statement 3 holds, we clearly have \(bHb^{−1}⊆H\). But, again using Statement 3, we also have \(b^{−1}Hb⊆H\); multiplying both sides of this equation on the left by \(b\) and on the right by \(b^{−1}\), we obtain \(H⊆bHb^{−1}\). Hence, since \(bHb^{−1}⊆H\) and \(H⊆bHb^{−1}\), those two sets are equal. Thus, Statement 2 is proven.
We now consider some examples and nonexamples of normal subgroups of groups.
Example \(\PageIndex{1}\)
- 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 \(7.2.1\) 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 \(2.8.9\).
Definition: Center
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.”)
Theorem \(\PageIndex{2}\)
Let \(G\) be a group. Then \(Z(G)\) is a subgroup of \(G\text{.}\)
- Proof
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Clearly, \(e∈Z(G)\). Next, let \(z,w∈Z(G)\). Then for all \(a∈G\),
\(a(zw)=(az)w=(za)w=z(aw)=z(wa)=(zw)a\),
so \(zw∈Z(G)\). Finally, \(az=za\) so, multiplying both sides on the left and right by \(z^{−1}\), we have \(z^{−1}a=az^{−1}\); thus, \(z^{−1}∈Z(G)\). Hence, \(Z(G)≤G\).
Theorem \(\PageIndex{3}\)
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 \(G\text{.}\)
- Proof
-
Let \(a∈G\). Then since every element of \(Z(G)\) commutes with every element of \(G\), every element of \(H\) commutes with every element of \(G\); so we must have \(aH=Ha\). Thus, \(H⊴G\).
The next definition is profoundly important for us.
Definition: Kernel
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 \(\phi\), \(\text{Ker} \;\phi\text{,}\) by
\begin{equation*} \text{Ker}\; \phi = \{k\in G\,:\,\phi(k)=e'\}. \end{equation*}
Theorem \(\PageIndex{4}\)
Let \(G\) and \(G'\) be groups and let \(\phi\) be a homomorphism from \(G\) to \(G'\text{.}\) Then \(\text{Ker}\; \phi\) is a normal subgroup of \(G\text{.}\)
- Proof
-
Let \(K=\text{Ker}\;ϕ\). We first show that \(K\) is a subgroup of \(G\). Clearly, the identity element of \(G\) is in \(K\), so \(K≠∅\). Next, let \(k,m∈K\). Then, letting \(e′\) denote the identity element of \(G′\), we have
\(ϕ(km^{−1})=ϕ(k)ϕ(m)^{−1}=e′(e′)^{−1}=e′\),
so \(km^{−1}∈K\). Thus, by the Two-Step Subgroup Test, we have that \(K\) is a subgroup of \(G\).
Next, let \(k∈K\) and let \(a∈G\). Then
\(ϕ(aka^{−1})=ϕ(a)ϕ(k)ϕ(a)^{−1}=ϕ(a)e′ϕ(a)^{−1}=ϕ(a)ϕ(a)^{−1}=e′\).
So \(aka^{−1}∈K\). Thus, \(K⊴G\).
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.
Example \(\PageIndex{2}\)
Let \(n\in \mathbb{Z}^+\text{.}\) Here is a rather elegant proof of the fact that \(SL(n,\mathbb{R})\) is a normal subgroup of \(GL(n,\mathbb{R})\text{:}\) Define \(\phi: GL(n,\mathbb{R}) \to \mathbb{R}^*\) by \(\phi(A)=\det A\text{.}\) Clearly, \(\phi\) is a homomorphism, and since the identity element of \(\mathbb{R}^*\) is \(1\),
\begin{equation*} \text{Ker} \;\phi=\{A\in GL(n,\mathbb{R})\,:\,\det A= 1\}=SL(n,\mathbb{R}). \end{equation*}
So \(SL(n,\mathbb{R})\unlhd GL(n,\mathbb{R})\text{.}\)