Normal Subgroups

# Normal Subgroups

Recall from the Left and Right Cosets of Subgroups page that if $(G, \cdot)$ is a group, $(H, \cdot)$ is a subgroup, and $g \in G$ then the left coset of $H$ with representative $g$ is defined to be the set:

(1)
\begin{align} \quad gH = \{ g \cdot h : h \in H \} \end{align}

Similarly, the right coset of $H$ with representative $g$ is defined to be the set:

(2)
\begin{align} \quad Hg = \{ h \cdot g : h \in H \} \end{align}

In general $gH$ need not equal $Hg$. That said, we can characterize special subgroups $(H, \cdot)$ of $(G, \cdot)$ for which their left and right cosets of $H$ with representative $g$ are equal for all $g \in G$.

 Definition: Let $(G, \cdot)$ be a group and $(H, \cdot)$ a subgroup. $(H, \cdot)$ is called a Normal Subgroup written $H \trianglelefteq G$ if $gH = Hg$ for all $g \in G$, that is, the left and right cosets of $H$ with representative $g$ are equal for all $g \in G$.

Immediately we can characterize a wide class of normal subgroups with the following theorem.

 Theorem 1: Let $(G, \cdot)$ be a group and $(H, \cdot)$ a subgroup. If $(G, \cdot)$ is abelian then $(H, \cdot)$ is a normal subgroup of $(G, \cdot)$.
• Proof: Let $(G, \cdot)$ be an abelian group. Then for all $a, b \in G$ we have that $a \cdot b = b \cdot a$. Let $g \in G$. Then $gH = \{ g \cdot h : h \in H \} = \{ h \cdot g : h \in H \} = Hg$ for all $g \in G$. So $(H, \cdot)$ is a normal subgroup of $(G, \cdot)$. $\blacksquare$

As a consequence of Theorem 1, the groups $(\mathbb{R}, +)$, $(\mathbb{Z}, +)$, $(\mathbb{Z}_n, +)$, $(\mathbb{Z}_p, \cdot)$, $(n\mathbb{Z}, +)$, etc… are all abelian groups so any subgroup of these groups is a normal subgroup.

Of course there exists nonabelian groups that have normal subgroups. For example, consider the group the symmetric group $(S_3, \circ)$ with $S_3 = \{ \epsilon, (12), (13), (23), (123), (132) \}$. This group is nonabelian. Let $H = \{ \epsilon, (123), (132) \}$. We claim that $(H, \circ)$ is a normal subgroup of $(S_3, \circ)$. Note that $\displaystyle{[S_3 : H] = \frac{\mid S_3 \mid}{\mid H \mid} = \frac{6}{3} = 2}$. The cosets of $H$ in $G$ are $\{ \epsilon, (123), (132) \}$ and $\{(12), (13), (23) \}$ and it's not hard to verify that $gH = Hg$ for all $g \in S_3$.