= Solution
The <kernel of a group homomorphism> is
$$
\boxed{K=\{g\in G:\Phi(g)=e_{G'}\}.}
$$
The homomorphism sends identity to identity and inverses to inverses. Thus $e_G\in K$, and if $k_1,k_2\in K$, then $\Phi(k_1k_2^{-1})=e_{G'}$, so $K$ is a subgroup by the <subgroup test>.
For any $g\in G$ and $k\in K$,
$$
\Phi(gkg^{-1})=\Phi(g)\Phi(k)\Phi(g)^{-1}=\Phi(g)e_{G'}\Phi(g)^{-1}=e_{G'}.
$$
Hence $gKg^{-1}\subseteq K$. Applying the same inclusion with $g^{-1}$ gives equality, proving
$$
\boxed{K\trianglelefteq G.}
$$
This normality argument does not require either group to be finite.
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