Normal subgroups

Let $G$ be an arbitrary group and suppose that $H_1$ and $H_2$ are subgroups of $G$. We say that $H_2$ is conjugate to $H_1$ if there exists an element $a \in G$ such that $H_2 = aH_1a^{- 1}$. It is easy to see, analogously to our consideration of conjugate elements, that this is an equivalence relation on the set of all subgroups of $G$. Thus one simply speaks of conjugate subgroups.

Analogous to the definition of the centralizer of an element $a \in G$, we define normalizer of the subgroup $H$ in $G$, denoted by $N_G(H)$ as follows:

\begin{displaymath} N_G(H)=\{a\in G\ \vert\ aH=Ha\}. \end{displaymath}

$N_G(H)$ is clearly a subgroup of $G$ and $H \subset N_G(H)$. (When it is clear which group we are working in, we may write $N(H)$ for $N_G(H)$.)

The following theorem is analogous to Theorem 4.3.4. The proof, which is the same except for notational changes, we leave to the reader as an exercise.

Theorem 6.1.1   Let $H \leq G$, $G$ a finite group. Then the number of subgroups of $G$ conjugate to $H$ (i.e., the order of the equivalence class containing $H$) is $[G: N_G(H)]$.

Next, we define the important notion of a normal subgroup.

Definition 6.1.2   A subgroup $H$ of a group $G$ is called a normal subgroup of $G$ if $aHa^{-1} = H$ for all $a \in G$. We denote this by $H \lhd G$.

Thus $H \lhd G$ if and only if $N_G(H) = G$. If $G$ is abelian, then every subgroup of $G$ is normal. For an arbitrary group $G$, it is clear that $G$ itself and the trivial subgroup $\{e\}$ are normal subgroups. In $S_3$ it is not hard to see that $A_3$ is normal. In fact $A_n \lhd S_n$ for all $n \in \mathbb{N}$. This follows as a special case of exercise 4 for this section. (It also follows by considering the parity of $gfg^{- 1}$ for $f \in A_n$ and $g\in S_n$.)

Suppose now that $N \lhd G$. If $aN$ and $bN$ are any two left cosets, then

\begin{displaymath} (aN) (bN) = a(Nb) N = abN^2 = abN \end{displaymath} (6.1)

which is a left coset. Here we have used the obvious fact that if $N$ is normal $bN = Nb$ for all $b \in G$. Thus if $N \lhd G$, we need not speak of left or right cosets with respect to $N$ because they are the same, so we can just talk simply of a coset.

The converse of this statement given in (6.1) above is also true and has been left as an exercise (see exercise 5 for this section). Namely, if $H \leq G$ is such that (for all $a,b \in G$) $aHbH = cH$ for some $c\in G$, i.e., the product of any two left cosets is a left coset, then $H \lhd G$.

For the sake of examples and references, we include here a table for the group $A_4$ of even permutations on the set $\{1, 2, 3, 4\}$. Let

\begin{displaymath} A_4=\{f_1,f_2,...,f_{12}\},\end{displaymath}

where

\begin{displaymath} \begin{array}{cccc} f_1=(1),& f_2=(1,2)(3,4),& f_3=(1,3)(2,4... ...& f_{10}=(2,4,3),& f_{11}=(2,3,4),& f_{12}=(1,2,4). \end{array}\end{displaymath}

The Alternating Group $A_4$.

$A_4$ $f_{1}$ $f_{2}$ $f_{3}$ $f_{4}$ $f_{5}$ $f_{6}$ $f_{7}$ $f_{8}$ $f_{9}$ $f_{10}$ $f_{11}$ $f_{12}$
$f_{1}$ $f_{1}$ $f_{2}$ $f_{3}$ $f_{4}$ $f_{5}$ $f_{6}$ $f_{7}$ $f_{8}$ $f_{9}$ $f_{10}$ $f_{11}$ $f_{12}$
$f_{2}$ $f_{2}$ $f_{1}$ $f_{4}$ $f_{3}$ $f_{6}$ $f_{5}$ $f_{8}$ $f_{7}$ $f_{10}$ $f_{9}$ $f_{12}$ $f_{11}$
$f_{3}$ $f_{3}$ $f_{4}$ $f_{1}$ $f_{2}$ $f_{7}$ $f_{8}$ $f_{5}$ $f_{6}$ $f_{11}$ $f_{12}$ $f_{9}$ $f_{10}$
$f_{4}$ $f_{4}$ $f_{3}$ $f_{2}$ $f_{1}$ $f_{8}$ $f_{7}$ $f_{6}$ $f_{5}$ $f_{12}$ $f_{11}$ $f_{10}$ $f_{9}$
$f_{5}$ $f_{5}$ $f_{8}$ $f_{6}$ $f_{7}$ $f_{9}$ $f_{12}$ $f_{10}$ $f_{11}$ $f_{1}$ $f_{4}$ $f_{2}$ $f_{3}$
$f_{6}$ $f_{6}$ $f_{7}$ $f_{5}$ $f_{8}$ $f_{10}$ $f_{11}$ $f_{9}$ $f_{12}$ $f_{2}$ $f_{3}$ $f_{1}$ $f_{4}$
$f_{7}$ $f_{7}$ $f_{6}$ $f_{8}$ $f_{5}$ $f_{11}$ $f_{10}$ $f_{12}$ $f_{9}$ $f_{3}$ $f_{2}$ $f_{4}$ $f_{1}$
$f_{8}$ $f_{8}$ $f_{5}$ $f_{7}$ $f_{6}$ $f_{12}$ $f_{9}$ $f_{11}$ $f_{10}$ $f_{4}$ $f_{1}$ $f_{3}$ $f_{2}$
$f_{9}$ $f_{9}$ $f_{11}$ $f_{12}$ $f_{10}$ $f_{1}$ $f_{3}$ $f_{4}$ $f_{2}$ $f_{5}$ $f_{7}$ $f_{8}$ $f_{6}$
$f_{10}$ $f_{10}$ $f_{12}$ $f_{11}$ $f_{9}$ $f_{2}$ $f_{4}$ $f_{3}$ $f_{1}$ $f_{6}$ $f_{8}$ $f_{7}$ $f_{5}$
$f_{11}$ $f_{11}$ $f_{9}$ $f_{10}$ $f_{12}$ $f_{3}$ $f_{1}$ $f_{2}$ $f_{4}$ $f_{7}$ $f_{5}$ $f_{6}$ $f_{8}$
$f_{12}$ $f_{12}$ $f_{10}$ $f_{9}$ $f_{11}$ $f_{4}$ $f_{2}$ $f_{1}$ $f_{3}$ $f_{8}$ $f_{6}$ $f_{5}$ $f_{7}$

Example 6.1.3   Let us consider the set $V_4 = \{f_1, f_2, f_3, f_4\}$. Referring to our table for $A_4$, it is easy to see that $V_4$ is a subgroup of $A_4$. (Recall from Exercise 4 in §4.2 that $V_4$ is called the Klein 4-group.) We claim that $V_4 \lhd A_4$. This can be shown by direct computation. (See exercise 7 for this section.) However, here we just note that since conjugacy preserves cycle structure, see Theorem 4.1.1, $gV_4g^{- 1} = V_4$ for all $g \in A_4$, thus $V_4 \lhd A_4$. $\Box$

We finally note that to show an arbitrary subgroup $N \lhd G$ it suffices to show $gng^{-1} \in N$ for all $g\in G$ because of the following result.

Proposition 6.1.4   Let $N \leq G$, $G$ a group. Then if $aNa^{- 1} \subset N$ for all $a \in G$, then $aNa^{- 1} = N$. In particular, $aNa^{- 1} \subset N$ for all $a \in G$ implies $N \lhd G$.

(See exercise 5(a) for this section for the proof.)


Subsections
David Joyner 2007-08-06