Solution
= Solution
A <group homomorphism> $f:G\to G'$ satisfies $f(gh)=f(g)f(h)$ for all $g,h\in G$. It consequently sends the <identity element> to the identity and inverses to inverses. Its <group homomorphism kernel> is
$$
\boxed{\ker f=\{g\in G:f(g)=e_{G'}\}}.
$$
A <normal subgroup> $H$ of $G$ is a <subgroup> invariant under <conjugation>:
$$
\boxed{gHg^{-1}=H\quad\text{for every }g\in G}.
$$
Equivalently, $gH=Hg$ for every $g$, so left and right <cosets> coincide.