= Solution
A <subgroup> $N\leq G$ is a <normal subgroup> if $gNg^{-1}=N$ for every $g\in G$. A <group homomorphism> $f:G\to\widetilde G$ satisfies $f(gh)=f(g)f(h)$ for every $g,h\in G$. In particular $f(e)=\widetilde e$ and $f(g^{-1})=f(g)^{-1}$. A <group representation> on a <vector space> $V$ is a <group homomorphism> $R:G\to\operatorname{GL}(V)$ into the <general linear group> of invertible <linear maps> on $V$.
For the <kernel of a group homomorphism>, put $K=\ker f=\{g:f(g)=\widetilde e\}$. It contains $e$, and if $k,\ell\in K$ then $f(k\ell^{-1})=f(k)f(\ell)^{-1}=\widetilde e$, so $K$ is a <subgroup>. For $g\in G$ and $k\in K$,
$$
f(gkg^{-1})=f(g)f(k)f(g)^{-1}=\widetilde e.
$$
Thus $gKg^{-1}\subseteq K$. 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> by
$$
\boxed{(gN)(hN)=(gh)N.}
$$
To check independence of representatives, replace $g,h$ by $gn,hm$, with $n,m\in N$. Then
$$
(gn)(hm)=gh(h^{-1}nh)m\in ghN,
$$
because 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 $N$.
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