3.6 Normal subgroups

In this section, \(G\) is a group with identity element \(e\).

Proposition 3.21

Suppose \(N\leq G\). Then the following are equivalent:

For all \(g\in G\) and \(n\in N\), \(gng^{-1}\in N\).
For all \(g\in G\), \(gNg^{-1} = N\).
For all \(g\in G\), \(gN = Ng\).
For all \(g\in G\), \(gNg^{-1} \subseteq N\).

Proof ▼

1 \(\Leftrightarrow \) 4 and 2 \(\Leftrightarrow \) 3 are very straightforward and left to the reader. And 2 \(\Rightarrow \) 4 is obvious. So it suffices to prove that 4 \(\Rightarrow \) 2.

To do this, suppose 4 holds, and take \(g\in G\). Then, applying 4 to \(g^{-1}\), we see that

\begin{equation} \label{ginvcont} g^{-1}N g\subseteq N. \end{equation}

3.22

But then,multiplying both sides of 3.22 on the left by \(g\), we see that

\[ Ng = g(g^{-1}Ng) \subseteq gN. \]

And, multiplying the last equation on both sides on the right by \(g^{-1}\), we see that \(N\subseteq gNg^{-1}\). So we get that \(N = gNg^{-1}\).

Definition 3.23

Suppose \(N\leq G\). We say that \(N\) is a normal subgroup of \(G\) and write \(N\unlhd G\) if the equivalent conditions of Proposition 3.21 hold.

Proposition 3.24

The trivial subgroup \(\{ e\} \) and \(G\) itself are both normal subgroups of \(G\).

Proof ▼

Obvious.

Proposition 3.25

If \(G\) is abelian, then every subgroup of \(G\) is normal.

Proof ▼

Obvious.

Definition 3.26

We say that \(G\) is simple if there are exactly two normal subgroups of \(G\).

Corollary 3.27

\(G\) is simple if and only if \(|G| {\gt} 1\) and \(G\) has no nontrivial proper normal subgoups.

Proof ▼

Follows from Propostion 3.24.

Example 3.28

Suppose \(G\) is a group of order \(p\) where \(p\) is a prime. Then \(G\) is simple.

Proof ▼

By Lagrange, \(G\) has no nontrivial proper subgroups. So, tautologically, it has no nontrivial proper normal subgoups.

Theorem 3.29

Suppose \(f:G\to H\) is a group homomorphism.

If \(K\unlhd H\), then \(f^{-1}(K)\unlhd G\). In particular, \(\ker f\unlhd G\).
If \(f\) is onto and \(N\unlhd G\), then \(f(N)\unlhd H\).

Proof ▼

1 Suppose \(K\unlhd G\). We already know that \(f^{-1}K \leq G\) by Theorem 3.17 2. So suppose \(g\in G\) and \(n\in f^{-1}K\). Then we have \(f(gng^{-1}) = f(g) f(n) f(g)^{-1} \in K\) by Proposition 3.21 1 as \(f(n)\in K\). So, by Proposition 3.21 1, \(f^{-1} K\unlhd G\).

2 Suppose \(f\) is onto and \(N\unlhd G\). Again, we already know that \(f(N)\leq H\). So, suppose \(y\in f(N)\) and \(h\in H\). We can write \(y = f(n)\) for some \(n\in N\), and, since \(f\) is onto, we can write \(h = f(g)\) for some \(g\in G\). Then \(gng^{-1}\in N\) as \(N\unlhd G\). So \(hyh^{-1} = f(gng^{-1}) \in f(N)\).

Example 3.30

Here’s an example of a subgroup, which is not normal. Take \(G = \operatorname{\mathbf{GL}}_2(\mathbb {R})\), and let \(T\) denote the set of all matrices in \(G\) of the form

\[ d(a,b) = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, \]

where, necessarily, \(ab\neq 0\). Then it is easy to see that \(T\leq G\).

Set

\[ g = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \]

and note that \(g\in G\) with

\[ g^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}. \]

But then

\[ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & b -a \\ 0 & b \end{pmatrix}. \]

Taking \((a,b) = (1,2)\), we see that \(d(a,b)\in T\), but \(gd(a,b)g^{-1}\not\in T\). So \(T\) is not normal in \(G\).

Using this, we can see why the assumption that \(f:G\to H\) is onto is necessary in Theorem 3.29 2: If \(i:T\to G\) is the inclusion homomorphism, then \(T \unlhd T\), but \(T = i(T)\) is not normal in \(G\).

Definition 3.31

Suppose \(H\leq G\). The normalizer of \(H\) in \(G\) is the subset \(N_G(H) = \{ g\in G: gH = Hg\} \).

Proposition 3.32

Suppose \(H\leq G\). Then \(H\leq N_G(H)\leq G\). Moreover, \(H\unlhd G\) if and only if \(N_G(H) = G\).

Proof ▼

Exercise!