If $\varphi :G \to H$ is an isomorphism, prove that $G$ is abelian if and only if $H$ is abelian. If
$\varphi :G \to H$ is a homomorphism, what additional conditions on $\varphi$ (if any) are sufficient to
ensure that if $G$ is abelian, then so is $H$?
My attempt :It is given that $G$ is abelian and $\varphi$ is a isomorphism,then for every element $h \in H$, there exists $g \in G $ such that $ h = \varphi(g)$. Then consider any two elements $h_1, h_2 \in H$ and let
$g_1, g_2 \in G$ be such that $h_1 = \varphi(g_1)$ and $h_2 = \varphi(g_2)$. We have $h_1h_2 = \varphi(g_1)\varphi(g_2) =\varphi(g_1g_2) = \varphi(g_2g_1) = \varphi(g_2)\varphi(g_1) = h_2h_1$. So $H$
is abelian
My doubt:If $\varphi :G \to H$ is a homomorphism,what additional conditions on $\varphi$ (if any) are sufficient to ensure that if $G$ is abelian, then so is $H$?
Best Answer
Note that in the proof you only used the fact that $\varphi$ is surjective. So this is sufficient.
In general, if $\varphi$ is any homomorphism and $G$ is abelian then it follows that the image $\varphi(G)$ is abelian as well. But if the image is not all $H$ (i.e if $\varphi$ is not surjective) then $H$ itself might not be abelian.