Question:
Suppose that $\alpha: G \to H$ is a surjective homomorphism of groups. Let $U$ be a subgroup of $H$ and prove the following:
The pre-image of $U$ under $\alpha$, .ie. $\{g \in G | g^{\alpha} \in U\}$, is a subgroup of $G$ containing $\ker(\alpha)$.
My answer:
Define $Y = \{g \in G | g^{\alpha} \in U\}$.
$1$. First showing $Y$ is a subgroup using the one-step subgroup test.
Let $g_1, g_2 \in Y$.
$(g_1g_2)^\alpha = g_1^\alpha g_2^\alpha \in U$
So, $g_2^\alpha \in U \implies (g_2^\alpha)^{-1} \in U \implies (g_2^{-1})^\alpha \in U \implies g_2^{-1} \in Y$
So we have $g_1g_2^{-1} \in Y$. And $Y$ is non-empty as $I_H \in U \implies I_G \in Y$ as homomorphisms preserve the identity element. Therefore $Y$ is a subgroup.
$2$. Now showing $\ker(\alpha) $ is contained in $Y$.
We have $I_H \in U$.
$\implies$ there exists $g_1, g_2,…,g_n \in Y$ such that $g_i^\alpha = I_H$, $i = 1,…,n$
So the set $\{g_1, g_2,…,g_n\} = \ker(\alpha)$ and hence $\ker(\alpha)$ is contained in $Y$.
How is that answer?
Best Answer
It looks good for the most part, though I think your justification for $\ker(\alpha)\subseteq Y$ is a bit shaky (the kernel need not be finite). You'll want to take an arbitrary member of $\ker(\alpha)$, say $g$, and show that $g\in Y$. Indeed, $g^\alpha=I_H\in U$, so by definition, $g\in Y$.