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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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Im a bit confused, is $g^\alpha = \alpha(g)$ ? – Stefan Nov 14 '12 at 21:20
Yes that is the notation our lecturer uses so I'm following it. I actually prefer $\alpha(g)$ myself. Which way is the most common way of writing it? – sonicboom Nov 14 '12 at 21:45
In my experience the functional notation is much more common. When I see $g^\alpha$ in a group theory context I expect it to mean $\alpha^{-1}g\alpha$ (or $\alpha g\alpha^{-1}$, depending on the writer’s convention for conjugation), where $\alpha$ is an element of the group. – Brian M. Scott Nov 14 '12 at 21:47
We don't even need the homomorphism to be surjective. – mez May 9 '14 at 9:03
up vote 2 down vote accepted

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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Cheers, I can see that its better to start out with taking the elements you want to use in the final conclusion than taking some other ones and 'transforming' them into what you need. – sonicboom Nov 14 '12 at 21:56
@sonicboom: Yes, and at least in these relatively straightforward arguments it’s usually best to go directly for what the relevant definitions require. – Brian M. Scott Nov 14 '12 at 21:58

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 $G$

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