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

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

• What do you mean $\phi(g)=G$?
– lulu
Mar 22 '17 at 18:43
• You mean $\phi(g)\in H$ and $\phi(h)\in H$? Mar 22 '17 at 18:44
• I said "Let $g,h\in G$". I am just learning so I was trying to see if it works. If not then I need some help trying to show it Mar 22 '17 at 18:44
• But, there is nothing to lose generality about. the group $G$ is abelian so $gh=hg$. The WLOG is nonsensical. Mar 22 '17 at 18:46
• Try writing your proof in words rather than in symbolism you're not familiar with. Then we can start looking at whether there are places where it would actually improve the clarity of your proof to start using symbols instead of words. Mar 22 '17 at 18:47

## 1 Answer

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: $x \cdot y = \phi(g) \phi(h) = \phi(gh) =\phi(hg) = \phi(h) \phi(g) = y \cdot x$.

• Oh I see what I did wrong here. These homomorphism proofs be confusing to me sometimes but I am starting to get a hang of it Mar 22 '17 at 18:53