$\alpha : G \to H$ 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?