Normal subgroup/Kernel/Introduction/Section

Definition

Let ${}G$ be a group, and let ${}H\subseteq G$ denote a subgroup. ${}H$ is called a normal subgroup if

${}xH=Hx\,$

holds for all ${}x\in G$, that is, if every left coset

of ${}x$ coincides with the right coset of ${}x$.

For a normal subgroup, it is not necessary to distinguish between left cosets and right cosets; we just talk about cosets. Instead of ${}xH$ or ${}Hx$, we write usually ${}[x]$. The equality ${}xH=Hx$ does not mean that ${}xh=hx$ for all ${}h\in H$; it only means that for every ${}h\in H$, there exists a ${}{\tilde {h}}\in H$ fulfilling ${}xh={\tilde {h}}x$.

Lemma

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. Then the following statements are equivalent.

1. ${}H$ is a normal subgroup of ${}G$.
2. We have ${}xhx^{-1}\in H$ for all ${}x\in G$ and ${}h\in H$.
3. ${}H$ is invariant under every inner automorphism of ${}G$.

Proof

(1) means for given ${}h\in H$ that we can write ${}xh={\tilde {h}}x$ with some ${}{\tilde {h}}\in H$. Multiplication by ${}x^{-1}$ from the right yields ${}xhx^{-1}={\tilde {h}}\in H$; therefore, ${}(2)$ holds. Reading this argument backwards gives the implication ${}(2)\Rightarrow (1)$. Moreover, ${}(2)$ is an explicit reformulation of ${}(3)$.

$\Box$

Example

We consider the permutation group ${}G=S_{3}$ for a set with three elements, that is, ${}S_{3}$ consists of all bijective mappings of the set ${}\{1,2,3\}$ to itself. The trivial group ${}\{\operatorname {id} \}$ and the whole group are normal subgroups. The subset ${}H=\{\operatorname {id} \,,\varphi \}$, where ${}\varphi$ is the element that swops ${}1$ and ${}2$ and fixes ${}3$, is a subgroup. However, it is not a normal subgroup. To show this, let ${}\psi$ denote the bijection that fixes ${}1$ and swops ${}2$ and ${}3$. The inverse of ${}\psi$ is ${}\psi$ itself. The conjugation ${}\psi \varphi \psi ^{-1}=\psi \varphi \psi$ is the mapping that sends ${}1$ to ${}3$, ${}2$ to ${}2$, and ${}3$ to ${}1$. This bijection does not belong to ${}H$.

Lemma

Let ${}G$ and ${}H$ be groups, and let

$\varphi \colon G\longrightarrow H$

be a group homomorphism. Then the kernel ${}\operatorname {kern} \varphi$ is a normal subgroup

in ${}G$.

Proof

By fact, we know that the kernel is a subgroup. We use fact. Hence, let ${}x\in G$ be arbitrary, and ${}h\in \operatorname {kern} \varphi$. Then

${}\varphi {\left(xhx^{-1}\right)}=\varphi (x)\varphi (h)\varphi {\left(x^{-1}\right)}=\varphi (x)e_{H}\varphi {\left(x^{-1}\right)}=\varphi (x)\varphi (x)^{-1}=e_{H}\,,$

therefore, ${}xhx^{-1}$ belongs to the kernel.

$\Box$