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May someone please verify my following proof?

For isomorphism $\phi: G\to H$ of the two groups $G$ and $H$, prove that $G$ is abelian iff $H$ is abelian.

Proof: (assume $G$ is abelian) Let $h_{1},h_{2}\in H$. Since $\phi$ is bijective, $\phi$ is onto. Then there exist $g_{1},g_{2}\in G$ such that $\phi (g_{1})=h_{1}$ and $\phi (g_{2})=h_{2}$ for all $h_{1},h_{2}\in H$. Then $h_{1}h_{2}=\phi (g_{1})\phi (g_{2})=\phi (g_{1}g_{2})$ (operation preserving) $=\phi (g_{2}g_{1})$ ($G$ is abelian) $=\phi (g_{2})\phi (g_{1})$ (operation preserving) $=h_{2}h_{1}$. Then $H$ is abelian.

(assume $H$ is abelian) Let $g_{1},g_{2}\in G$. Since $\phi$ is bijective, the inverse $\phi ^{-1}:H\to G$ exists and is onto. Then there exist $h_{1},h_{2}\in H$ such that $\phi ^{-1}(h_{1})=g_{1}$ and $\phi ^{-1}(h_{2})=g_{2}$ for all $g_{1},g_{2}\in G$. Then $g_{1}g_{2}=\phi ^{-1} (h_{1})\phi ^{-1}(h_{2})=\phi ^{-1}(h_{1}h_{2})$ (operation preserving) $=\phi ^{-1}(h_{2}h_{1})$ ($H$ is abelian) $=\phi ^{-1}(h_{2})\phi ^{-1}(h_{1})$ (operation preserving) $=g_{2}g_{1}$. Then $G$ is abelian. $\square$

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    Related. In some sense there is really nothing to prove. https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker Mar 02 '18 at 22:07
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    It is correct. For the second part you could use the fact that if $\phi$ is an isomorphism then so is $\phi^{-1}$. – juan arroyo Mar 02 '18 at 22:07

2 Answers2

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It is correct but, in the first paragraph, you shouldn't have written “for all $h_1,h_2\in H$”. The elements $h_1$ and $h_2$ were given at the beginning of the proof.

And it is a waste of time assirting that $\phi^{-1}$ is onto. Every inverse has that property.

José Carlos Santos
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You should change all the $\phi$ to $\phi^{-1}$ starting the second line of the second paragraph. Also, I agree with Jose on that you should drop "for all $h_{1},h_{2}\in H$" and "for all $g_{1},g_{2}\in G$" because you already picked those elements at the beginning so it is already fixed. It also sounds like $\phi(g_{1})=h_{1}$ for all $h_{1}\in H$, i.e., $\phi(g_{1})$ equals every element of $H$, which is wrong.

Delong
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  • Opps, I totally forgot to write the inverse notation. Thanks for bringing that up. –  Mar 02 '18 at 22:54