Question:
Suppose that $\alpha: G \to H$ is a surjective homomorphism of groups. Let $U$ be a subgroup of $H$ and prove the following:
If $U$ is normal in $H$, the pre-image of $U$ is normal in $G$.
My answer:
Define $Y = \{g \in G | g^{\alpha} \in U\}$.
$U$ is normal in $H$ so $h^{-1}uh \in U$ for $h \in H, u \in U$.
Now $h = g^\alpha$ and $u = y^\alpha$ for $g \in G, y \in Y$.
So we have $(g^\alpha)^{-1}(y^\alpha)(g^\alpha) \in U$
$\implies (g^{-1}yg)^\alpha \in U$
$\implies g^{-1}yg \in Y$
So $Y$ is normal in $G$.
Is that correct?