Have $\alpha:G \to H$. If a subgroup $U$ is normal in $H$ prove the pre-image of $U$ is normal in $G$.

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 $G$.

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?

-
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 at 9:03

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.
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 $G$