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

  • 1
    $\begingroup$ What do you mean $\phi(g)=G$? $\endgroup$
    – lulu
    Mar 22, 2017 at 18:43
  • $\begingroup$ You mean $\phi(g)\in H$ and $\phi(h)\in H$? $\endgroup$
    – user296113
    Mar 22, 2017 at 18:44
  • $\begingroup$ 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 $\endgroup$
    – Lady T
    Mar 22, 2017 at 18:44
  • 1
    $\begingroup$ But, there is nothing to lose generality about. the group $G$ is abelian so $gh=hg$. The WLOG is nonsensical. $\endgroup$
    – David Hill
    Mar 22, 2017 at 18:46
  • 4
    $\begingroup$ 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. $\endgroup$ Mar 22, 2017 at 18:47

1 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 $.

  • $\begingroup$ 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 $\endgroup$
    – Lady T
    Mar 22, 2017 at 18:53

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