10.Normal subgroup

Definition

Let $H$ be a subgroup of $G$ (i.e., $H\le G$). If for all $a\in G$, $aH=Ha$, then $H$ is called a normal subgroup of $G$ (or an invariant subgroup), denoted as $H⊴G$.

In this case, the left and right cosets are simply called cosets.

The two trivial subgroups of $G$, $\{e\}$ and $G$ itself, are normal subgroups.

If $H$ is a nontrivial proper normal subgroup of $G$, it can be denoted as $H⊲G$.

If $G$ is an abelian group, then any subgroup of $G$ is a normal subgroup.

All subgroups of the groups $<\mathbb{Z},+>$, $<{\mathbb{Z}}_n,+>$, and $<{\mathbb{Z}}_n^{\ast },\otimes >$ are normal subgroups.

Normal subgroups are cosets equal on the basis of subgroups.

Criterion

Based on definition

Let $G=\{\left(\begin{array}{cc}r& s\\ 0& 1\end{array}\right)\mid r,s\in \mathbb{Q},r\ne 0\}$, $H=\{\left(\begin{array}{cc}1& s\\ 0& 1\end{array}\right)\mid s\in \mathbb{Q}\}$, then $H$ is a subgroup of $G$. Prove that $H$ is a normal subgroup of $G$.

Proof: For $\mathrm{\forall }a=\left(\begin{array}{cc}r& t\\ 0& 1\end{array}\right)\in G$, where $r,t\in \mathbb{Q},r\ne 0$, we have

$$aH=\{\left(\begin{array}{cc}r& r{s}_1+t\\ 0& 1\end{array}\right)\mid s_1\in \mathbb{Q}\},$$$$Ha=\{\left(\begin{array}{cc}r& s_2+t\\ 0& 1\end{array}\right)\mid s_2\in \mathbb{Q}\}.$$

Since $r{s}_1\in \mathbb{Q}$, clearly $aH\subseteq Ha$.

Conversely, for $s_2+t$, since $r\ne 0$, choose $s_1=\frac{s_2+t}{r}$, then $r{s}_1+t=s_2+t$, so $Ha\subseteq aH$.

Therefore, $aH=Ha$, $H⊴G$.

Necessary and sufficient condition

Let $H$ be a subgroup of $G$. A necessary and sufficient condition for $H$ to be a normal subgroup of $G$ is:

For $\mathrm{\forall }a\in G$, $h\in H$, there is $ah{a}^{-1}\in H$.

Proof:

Necessity: If $H$ is a normal subgroup of $G$, then for $\mathrm{\forall }a\in G$, $h\in H$, there exists $h_1\in H$, such that $ah=h_{1}a$, thus $ah{a}^{-1}=h_1\in H$, so $ah{a}^{-1}\in H$.

Sufficiency: For $\mathrm{\forall }a\in G$, $h\in H$, since $ah{a}^{-1}\in H$, there exists $h_1\in H$, such that $ah=h_{1}a$, thus $aH\subseteq Ha$. Also, $ha\in Ha$, for $\mathrm{\forall }a\in G$, $h\in H$, $a^{-1}\in G$, then $a^{-1}ha\in H$, so there exists $h_2\in H$, such that $ha=a{h}_2$, thus $Ha\subseteq aH$. Hence, for $\mathrm{\forall }a\in G$, it holds that $aH=Ha$.

Therefore, $H$ is a normal subgroup of $G$.

A homomorphic kernel is a normal subgroup

Let $(G,\ast )$ and $(G_1,\circ )$ be groups, $e,e^{\prime }$ be the identities of $G$ and $G_1$, respectively. If $f:G\to G_1$ is a homomorphism, then the kernel $\text{Ker}f$ is a normal subgroup of $G$.

Proof:

(1) Prove $\text{Ker}f$ is a subgroup.

Since $f$ is a homomorphism, we have $f(e)=e^{\prime }$. Hence $e\in \text{Ker}f$, and $\text{Ker}f$ is a non-empty subset of $G$. For $\mathrm{\forall }a,b\in \text{Ker}f$, by the definition of the kernel we have $f(a)=e^{\prime }$, $f(b)=e^{\prime }$. Since $f$ is a homomorphism, we have $f(a\ast b^{-1})=f(a)\circ f(b^{-1})=f(a)\circ (f(b){)}^{-1}=e^{\prime }\circ (e^{\prime }{)}^{-1}=e^{\prime }$, which means $a\ast b^{-1}\in \text{Ker}f$. Therefore, $\text{Ker}f\subseteq ⟨G,\ast ⟩$ is a subgroup.

(2) Prove $\text{Ker}f$ is a normal subgroup.

For $\mathrm{\forall }n\in \text{Ker}f$, $\mathrm{\forall }a\in G$, we have $f(a\ast n\ast a^{-1})=f(a)\circ f(n)\circ f(a^{-1})=f(a)\circ e^{\prime }\circ (f(a){)}^{-1}=e^{\prime }$, thus $a\ast n\ast a^{-1}\in \text{Ker}f$.

From (1) and (2), we know that $\text{Ker}f$ is a normal subgroup of $G$.