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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:

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?

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    We don't even need the homomorphism to be surjective.2014-05-09

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It’s not entirely correct, even if $g^\alpha$ is just an unusual way of writing $\alpha(g)$: $\alpha$ need not be injective. Starting with $Y=\alpha^{-1}[U]$ is fine. Now you want to prove that $Y$ is normal in $G$, so don’t start over in $H$: you want to show that for each $g\in G$ and $y\in Y$, $g^{-1}yg\in Y$, so start with an arbitrary $g\in G$ and $y\in Y$. Now $\alpha(g^{-1}yg)=\alpha(g^{-1})\alpha(y)\alpha(g)\in U\;,$ since $\alpha(y)\in U$ and $U$ is normal in $H$, so by definition $g^{-1}yg\in Y$, which is exactly what you need to prove.

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    @sonicboom: Yes, and at least in these relatively straightforward arguments it’s usually best to go directly for what the relevant definitions require.2012-11-14
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if $U$ is a normal subgroup of $H$ then $\exists \beta, H'$ such that $\beta:H \rightarrow H'$ with kernel $U$.

so $\beta \alpha:G \rightarrow H'$ has kernel $U$ and therefore $U$ must be normal in $G$