TITLE: Showing that if a subgroup is normal, it's surjective homomorphic image is normal
QUESTION [6 upvotes]: $\alpha : G \to H$ is a surjective homomorphism. And $U \subset G$ is a subgroup of $G$. Verfiy the claim -
The image of $U$, ie $\alpha(U)$, is a subgroup of $H$, and if $U$ is normal in $G$, then $\alpha(U)$ is normal in $H$.
Answer:
Firstly, do I have to show $\alpha(U)$, is a subgroup of $H$ or is that statement just a statement of fact as part of the question?
Anyway here is what I have done..taking it as a given that $\alpha(U)$, is a subgroup of $H$ -
As $U$ is normal we have 
$U = gUg^{-1}$
$\alpha(U) = \alpha(gUg^1) =$ {applying homomorphic mapping into H} = $\alpha(g)\alpha(U)\alpha(g^{-1})$
Is that correct? I have a feeling I should take the $-1$ exponent outside the bracket as an extra final step or is that superfluous?

REPLY [2 votes]: To show that $\alpha(U)$ is a normal subgroup, you need to prove that $\alpha(U) = x \alpha(U) x^{-1}$ for all $x \in H$.  But any $x \in H$ can be written in the form $\alpha(g)$ for $g \in G$ since $\alpha$ is surjective.  Thus, you need only prove that $\alpha(U) = \alpha(g) \alpha(U) \alpha(g)^{-1}$ for all $g \in G$.  Note the $-1$ exponent is "on the outside", so you really do need to take that last step as you suspected.
And yes, as written it is meant that you should prove that $\alpha(U)$ is a subgroup.  The proof requires you to carefully work through the definition/criterion for being a subgroup, but nothing beyond that.
Also, +1 for showing your reasoning and clearly indicating what you are unsure of.