Proving $\phi(G)$ is Abelian if $\phi$: $G \to H$ is a group homomorphism

Question

Question: If $\phi$ : $G \to H$ is a group homomorphism and $G$ is abelian, Prove that $\phi(G)$ is also abelian

Here is my attempt:
Let $g$,$h\in G$
then $\phi(g)$=$G$ and $\phi(h)$=$G$
$\Rightarrow$ $\phi(gh)$=$\phi(g)\phi(h)$
$\Rightarrow$$g\cdot h$
By using the definition of Abelian, $x\cdot y$ = $y\cdot x$ $\forall$ $x,y\in G$
$g\cdot h$ = $h\cdot g$ $\forall$ $g,h\in G$
$\Rightarrow$ $\phi(gh)$=$\phi(hg)$

Am I right? If not then I need help

Answer 1

I think something went wrong with your notation but you seem to have the right idea. Let $x,y \in \phi (G) $, say $x = \phi (g), y = \phi (h) $ for $g,h \in G $. To see commutativity, you can compute: $x \cdot y = \phi(g) \phi(h) = \phi(gh) =\phi(hg) = \phi(h) \phi(g) = y \cdot x $.