 Solution

A subgroup $N\le G$ is a normal subgroup if $gN{g}^{-1}=N$ for every $g\in G$. A group homomorphism $f:G\to \tilde{G}$ satisfies $f(gh)=f(g)f(h)$ for every $g,h\in G$. In particular $f(e)=\tilde{e}$ and $f(g^{-1})=f(g{)}^{-1}$. A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$ into the general linear group of invertible linear maps on $V$.

For the kernel of a group homomorphism, put $K=\ker f=\{g:f(g)=\tilde{e}\}$. It contains $e$, and if $k,\ell \in K$ then $f(k{\ell }^{-1})=f(k)f(\ell {)}^{-1}=\tilde{e}$, so $K$ is a subgroup. For $g\in G$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Applying this inclusion with $g^{-1}$ gives the reverse inclusion, proving that $\ker f$ is a normal subgroup.

For a normal subgroup $N$, define multiplication of left cosets byTo check independence of representatives, replace $g,h$ by $gn,hm$, with $n,m\in N$. Thenbecause normality gives $h^{-1}nh\in N$. Hence the operation is well-defined. Associativity follows from $(gh)k=g(hk)$ in $G$; the identity is $N=eN$; and the inverse of $gN$ is $g^{-1}N$. These verify the group axioms, giving the quotient group $G/N$. The map $g\mapsto gN$ is a surjective group homomorphism with kernel of a group homomorphism equal to $N$.