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 ${\mathrm{\ast }}_G$, ${\mathrm{\ast }}_H$ and identity elements $e_G$ and $e_H$, respectively) and let $\mathrm{\Phi }\mathrm{:}G\mathrm{\to }H$ be a group homomorphism. Then, the kernel of $\mathrm{\Phi }$, i.e.

$$\mathrm{Ker}(\mathrm{\Phi })=\{g\in G:\mathrm{\Phi }(g)=e_H\},$$

is a normal subgroup of $G$.

Proof.

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

$\mathrm{\Phi }(g{\ast }_{G}k{\ast }_G{g}^{-1})$ $=$ $\mathrm{\Phi }(g){\ast }_H\mathrm{\Phi }(k){\ast }_H\mathrm{\Phi }(g^{-1})$

$=$ $\mathrm{\Phi }(g){\ast }_H(e_H){\ast }_H\mathrm{\Phi }(g^{-1})$

$=$ $\mathrm{\Phi }(g){\ast }_H\mathrm{\Phi }(g^{-1})$

$=$ $\mathrm{\Phi }(g){\ast }_H\mathrm{\Phi }{(g)}^{-1}$

$=$ $e_H,$

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

Conversely:

Lemma 2.

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

Proof.

Simply set $H$ equal to the quotient group $G/K$ and define $\mathrm{\Phi }:G\to G/K$ to be the natural projection from $G$ to $G/K$ (i.e. $\mathrm{\Phi }$ sends $g\in G$ to the coset $gK$). Then it is clear that the kernel of $\mathrm{\Phi }$ 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 $\mathrm{SL}(n,F)$ is normal inside the general linear group $\mathrm{GL}(n,F)$, for all $n\ge 1$. By the lemmas, it suffices to construct a homomorphism of $\mathrm{GL}(n,F)$ with $\mathrm{SL}(n,F)$ as kernel. The determinant of matrices is the homomorphism we are looking for. Indeed:

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

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

Title	the kernel of a group homomorphism is a normal subgroup
Canonical name	TheKernelOfAGroupHomomorphismIsANormalSubgroup
Date of creation	2013-03-22 17:20:34
Last modified on	2013-03-22 17:20:34
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	8
Author	alozano (2414)
Entry type	Theorem
Classification	msc 20A05
Related topic	KernelOfAGroupHomomorphism
Related topic	NaturalProjection