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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?

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    Note that one doesn't actually need the surjectivity hypothesis.2015-05-21

1 Answers 1

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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, $g\in Y$.

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    Ok I get it now, cheers. I am putting down - Let $g \in ker(\alpha)$. As $g^\alpha = I_H \in U \implies g^\alpha \in Y$ and hence $ker(\alpha) \in Y$.2012-11-14