= Solution
A <group homomorphism> satisfies $\varphi(ab)=\varphi(a)\varphi(b)$ for every $a,b\in G$. Its <kernel of a group homomorphism> is $K=\{g:\varphi(g)=e_H\}$. The homomorphism equation gives $\varphi(e_G)=e_H$ by cancellation, and $\varphi(g^{-1})=\varphi(g)^{-1}$. Therefore $K$ contains the identity and is closed under multiplication and inverses, so it is a <subgroup>.
For $k\in K$ and $x\in G$,
$$
\varphi(x^{-1}kx)=\varphi(x)^{-1}e_H\varphi(x)=e_H.
$$
Thus conjugation carries $K$ into itself, and applying the inverse conjugation gives equality: $K$ is a <normal subgroup>. If $K=\{e,\xi\}$, conjugation cannot send $\xi$ to the identity because it is injective. Consequently
$$
\boxed{x^{-1}\xi x=\xi\quad\text{for every }x\in G.}
$$
This proves directly the <normal subgroup of order two is central> principle for kernels, including the kernel facts needed below.
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