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Could someone please verify my following proof?

For isomorphism $\phi:G\to H$, show that $ab=ba$ for $a,b\in G$ iff $\phi(a)\phi(b)=\phi(b)\phi(a)$.

Proof:

Let $ab=ba$ for all $a,b\in G$. Then $G$ is abelian. Then $H$ is abelian. Since $\phi$ is bijective, $H=\phi(G)=\left \{ \phi(g) : g\in G \right \}$. Therefore, $\phi(a)\phi(b)=\phi(b)\phi(a)$ for all $\phi(a),\phi(b)\in H$.

Let $\phi(a)\phi(b)=\phi(b)\phi(a)$ for all $\phi(a),\phi(b)\in H$. Since $\phi$ is bijective, $H=\phi(G)=\left \{ \phi(g) : g\in G \right \}$. Then $H$ is abelian. Then $G$ is abelian. Then $ab=ba$ for all $a,b\in G$.

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    $\begingroup$ This is the right idea, a little clumsily written. But there's a sense in which it is too much work. You shouldn't have to make any argument at all, except to practice arguments. See math.stackexchange.com/questions/2039702/… $\endgroup$ Mar 6 '18 at 21:12
  • $\begingroup$ Are you sure the intent is for all $a,b \in G$ and not for any? $\endgroup$
    – gt6989b
    Mar 6 '18 at 21:13
  • $\begingroup$ @gt6989b I am not sure what you mean. Don't the universal quantifiers "for all" and "for any" mean the same thing? $\endgroup$
    – user482939
    Mar 6 '18 at 21:39
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    $\begingroup$ I mean you don't need to assume the whole group is Abelian, just that these specific $a,b \in G$ commute $\endgroup$
    – gt6989b
    Mar 6 '18 at 22:35
  • $\begingroup$ @gt6989b I understand what you mean now...thanks, I will rewrite my proof without the abelian part! $\endgroup$
    – user482939
    Mar 7 '18 at 0:26
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If $\phi:G\to H$ is any homomorphism, and $a,b\in G$ are a particular commuting pair of elements: $ab=ba$ (we don't need to assume that the whole $G$ is Abelian), then we already have $$\phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a)\,.$$ If $\phi$ is an isomorphism, the converse also holds for the particular commuting pair $\phi(a),\phi(b)$: specifically, we can apply the same argument for the homomorphism $\phi^{-1}$.

Of course, your weaker interpretation of (the quantifiers in) the statement is also valid, and your proof is correct.

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  • $\begingroup$ I see now what you mean about the quantifiers. I understand now why my proof wasn't a good approach...thanks! $\endgroup$
    – user482939
    Mar 7 '18 at 0:26

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