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. $$\Ker(\Phi)=\{ g\in G : \Phi(g)=e_H\},$$ 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 $k\in \Ker(\Phi)$ Then, $\Phi(k)=e_H$ by definition and: \begin{eqnarray*} \Phi(g \, \ast_G \, k \, \ast_G \, g^{-1}) &=& \Phi(g)\ast_H \Phi(k) \ast_H \Phi(g^{-1})\\ & = & \Phi(g)\ast_H ( e_H ) \ast_H \Phi(g^{-1})\\ & = & \Phi(g) \ast_H \Phi(g^{-1}) \\ & = & \Phi(g) \ast_H \Phi(g)^{-1}\\ & = & e_H, \end{eqnarray*}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 $k\in \Ker(\Phi)$ were arbitrary, it follows that $\Ker(\Phi)$ 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 $K$

Example 1   Let $F$ be a field. Let us prove that the special linear group $\SL(n,F)$ is normal inside the general linear group $\GL(n,F)$ for all $n\geq 1$ By the lemmas, it suffices to construct a homomorphism of $\GL(n,F)$ with $\SL(n,F)$ as kernel. The determinant of matrices is the homomorphism we are looking for. Indeed: $$\det : \GL(n,F) \to F^\times$$ is a group homomorphism from $\GL(n,F)$ to the multiplicative group $F^\times$ and, by definition, the kernel is precisely $\SL(n,F)$ i.e. the matrices with determinant $=1$ Hence, $\SL(n,F)$ is normal in $\GL(n,F)$