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Could someone please verify whether my proof is correct or is missing anything important?

Prove that for homomorphism $\phi:G\to H$ if $G_{1}\leq G$ is abelian, then $\phi(G_{1})$ is abelian.

Let $G_{1}$ be abelian. Then $ab=ba$ $ \forall a,b\in G_{1}$. Since $\phi(G_{1})=\left \{ \phi(a):a\in G_{1} \right \}$, by definition of homomorphisms, $\phi(a)\phi(b)=\phi(ab)=\phi(ba)$ ($G_{1}$ is abelian) $=\phi(b)\phi(a)$. Then $\phi(G_{1})$ is abelian.

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Yes, it is correct. I suggest that you start with two elements $x,y\in\phi(G_1)$. You want to prove that $xy=yx$. But, since, $x,y\in\phi(G_1)$, there are $a,b\in G_1$ such that $\phi(a)=x$ and $\phi(b)=y$. Then\begin{align}xy&=\phi(a)\phi(b)\\&=\phi(ab)\\&=\phi(ba)\\&=\phi(b)\phi(a)\\&=yx.\end{align}

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  • $\begingroup$ I understand what you mean. Thank you for taking the time to write it out as well. Is your way more correct or is my shortcut fine as well? $\endgroup$
    – user482939
    Mar 15 '18 at 19:22
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    $\begingroup$ @numericalorange What you did is correct. I just fell that the way I did is more natural but it's really just a matter of taste. $\endgroup$ Mar 15 '18 at 19:23

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