# the kernel of a group homomorphism is a normal subgroup

In this entry we show the following simple lemma:

###### Lemma 1.

Let $G$ and $H$ be groups (with group operations $\ast_{G}$, $\ast_{H}$ and identity elements $e_{G}$ and $e_{H}$, respectively) and let $\Phi:G\to H$ be a group homomorphism. Then, the kernel of $\Phi$, i.e.

 $\operatorname{Ker}(\Phi)=\{g\in G:\Phi(g)=e_{H}\},$

is a normal subgroup of $G$.

###### Proof.

Let $G,H$ and $\Phi$ be as in the statement of the lemma and let $g\in G$ and $k\in\operatorname{Ker}(\Phi)$. Then, $\Phi(k)=e_{H}$ by definition and:

 $\displaystyle\Phi(g\,\ast_{G}\,k\,\ast_{G}\,g^{-1})$ $\displaystyle=$ $\displaystyle\Phi(g)\ast_{H}\Phi(k)\ast_{H}\Phi(g^{-1})$ $\displaystyle=$ $\displaystyle\Phi(g)\ast_{H}(e_{H})\ast_{H}\Phi(g^{-1})$ $\displaystyle=$ $\displaystyle\Phi(g)\ast_{H}\Phi(g^{-1})$ $\displaystyle=$ $\displaystyle\Phi(g)\ast_{H}\Phi(g)^{-1}$ $\displaystyle=$ $\displaystyle e_{H},$

where we have used several times the properties of group homomorphisms and the properties of the identity element $e_{H}$. Thus, $\Phi(gkg^{-1})=e_{H}$ and $gkg^{-1}\in G$ is also an element of the kernel of $\Phi$. Since $g\in G$ and $k\in\operatorname{Ker}(\Phi)$ were arbitrary, it follows that $\operatorname{Ker}(\Phi)$ is normal in $G$. ∎

###### Lemma 2.

Let $G$ be a group and let $K$ be a normal subgroup of $G$. Then there exists a group homomorphism $\Phi:G\to H$, for some group $H$, such that the kernel of $\Phi$ is precisely $K$.

###### Proof.

Simply set $H$ equal to the quotient group $G/K$ and define $\Phi:G\to G/K$ to be the natural projection from $G$ to $G/K$ (i.e. $\Phi$ sends $g\in G$ to the coset $gK$). Then it is clear that the kernel of $\Phi$ is precisely formed by those elements of $K$. ∎

Although the first lemma is very simple, it is very useful when one tries to prove that a subgroup is normal.

###### Example.

Let $F$ be a field. Let us prove that the special linear group $\operatorname{SL}(n,F)$ is normal inside the general linear group $\operatorname{GL}(n,F)$, for all $n\geq 1$. By the lemmas, it suffices to construct a homomorphism of $\operatorname{GL}(n,F)$ with $\operatorname{SL}(n,F)$ as kernel. The determinant of matrices is the homomorphism we are looking for. Indeed:

 $\det:\operatorname{GL}(n,F)\to F^{\times}$

is a group homomorphism from $\operatorname{GL}(n,F)$ to the multiplicative group $F^{\times}$ and, by definition, the kernel is precisely $\operatorname{SL}(n,F)$, i.e. the matrices with determinant $=1$. Hence, $\operatorname{SL}(n,F)$ is normal in $\operatorname{GL}(n,F)$.

Title the kernel of a group homomorphism is a normal subgroup TheKernelOfAGroupHomomorphismIsANormalSubgroup 2013-03-22 17:20:34 2013-03-22 17:20:34 alozano (2414) alozano (2414) 8 alozano (2414) Theorem msc 20A05 KernelOfAGroupHomomorphism NaturalProjection