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# A subgroup is normal if and only if it is the kernel of a homomorphism.

We are going tho show that the kernel of a group homomorphism is a normal subgroup. Next, we prove that every normal subgroup is the kernel of a group homomorphism.

## Kernel of a group homomorphism is a normal subgroup

Let $\varphi: G \rightarrow H$ a group homomorphism and $\operatorname{ker} \varphi=\{g \in G: \varphi(g)=e_H\}$

$\ker \varphi$ is a subgroup of $G$.

1/ If $a, b \in \operatorname{ker}(\varphi)$ then the product $a b \in \operatorname{ker}(\varphi)$:

$$\varphi(a b)=\varphi(a) \varphi(b)=e_H \cdot e_H =e_H$$

2/ $e_G \in \operatorname{ker}(\varphi)$ ie. $\varphi(e_G)=e_H$:
For all $a \in G, a \cdot e_G=a=e_G \cdot a$. Since $\varphi$ a homomorphism:

\begin{aligned} \varphi(a \cdot e_G) &=\varphi(a)=\varphi(e_G \cdot a) \\ \varphi(a) \varphi(e_G) &=\varphi(a)=\varphi(e_G) \varphi(a)\\ h \varphi(e_G) &=h=\varphi(e_G) h\\ \end{aligned}

with $\varphi(a)=h$. Every factor are elements of the group $H$ . By definition and uniqueness of identity element of $H$:

$$\varphi(e_G)=e_H$$

3/ If $a \in \operatorname{ker}(\varphi)$ then $a^{-1} \in \operatorname{ker}(\varphi)$ ie. $\varphi\left(a^{-1}\right)=e_H$

$a$ and $a^{-1}$ are inverses in $G \Longleftrightarrow a a^{-1}=e_G=a^{-1} a$.

$$\begin{array}{c} \varphi\left(a a^{-1}\right)=\varphi(e_G)=\varphi\left(a^{-1} a\right) \\ \varphi(a) \varphi\left(a^{-1}\right)=\varphi(e_G)=\varphi\left(a^{-1}\right) \varphi(a) \end{array}$$

According to the previous proof $\varphi(e_G)=e_H$ and since $\varphi(a)=e_H$:

$$e_H \cdot \varphi\left(a^{-1}\right)=e_H=\varphi\left(a^{-1}\right) \cdot e_H$$

By definition of the identity of $H$, we have $\varphi\left(a^{-1}\right)=e_H$

$\ker \varphi$ is a normal subgroup of $G$.

If $a \in G, b \in \operatorname{ker} \varphi$ then $a b a^{-1} \in \ker \varphi$

$$\varphi\left(a b a^{-1}\right)=\varphi(a) \cdot \varphi(b) \cdot \varphi \left(a^{-1}\right)=\varphi(a) \cdot e_h \cdot \varphi(a)^{-1}=\varphi(a) \cdot \varphi(a)^{-1}=e_h$$

## Every normal subgroup is the kernel of a group homomorphism

Let $N$ a normal subgroup of $G$. We must construct a homomorphism $\varphi$ and a group $H$ such $N=\ker \varphi$.

$$\forall a\in G,b\in N, \quad a b a^{-1} \in N \iff \forall a\in G, \quad aN=Na$$

Since $N$ is a normal subgroup of $G$, we can define the group quotient:

$$H= G/N = \{aN : a \in G\} = \{Na : a \in G\}$$

with the operation $(a N, b N) \mapsto a b N$. Let’s verify that this operation is well defined independently of the cosets representation of $aN$ and $bN$. Suppose now we have $a_{1} \in a N$ and $b_{1} \in bN$, such

$$\left(a_{1} N, b_{1} N\right) \mapsto a b N$$

Since $a_1 \in aN, b_{1} \in bN$, there exists $n_1, n_2\in N$ such $a_{1}=a n_1, b_1=b n_2$ :

$$a_{1} N=a n_1 N=a N, b_{1} N=b n_2 N=b N$$

Then:

$$\left(a_{1} N\right)\left(b_{1} N\right)=(a N)\left(bN\right)=a b N$$

Closure: $\left(a N\right)\left(b N\right)=a\left(N b\right) N=a\left(bN\right) N=a b NN=abN \in H$
Associativity: follows from definition of $G$.
Identity: $e_H=e_GN=N$ is the identity in $H$.
Inverse: $a^{-1} N$ is the inverse of $a N$ in $H$: $(a N)\left(a^{-1} N\right)=\left(a a^{-1}\right) N=N$

Now we define the map $\varphi: G \rightarrow H$ by $\varphi(a)=a N$. Then $\varphi$ is a homomorphism since:

$$\varphi(a b)=a b N=(a N)(b N)=\varphi(a) \varphi(b)$$

Then :

$$\operatorname{ker} \varphi=\{a \in G: \varphi(a)=a N=N\}=\{a \in G: a \in N\}= N$$