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I would like to know if my proof is correct. Specifically, I would like you to check that surjectivity is needed for proving the first part and injectivity is needed for proving the second part.


Problem If $\phi: G \to H$ is an isomorphism, prove that $G$ is abelian if and only if $H$ is abelian.

Solution

Let $\phi: G \to H$ be an isomorphism.

  • If $G$ is abelian, we have $xy = yx$, for any $x,y \in G$. Consider any two element in $H$, say $c$ and $d$. Since $\phi$ is an isomorphism (more specifically a surjection), we have $c = \phi(a)$ and $d=\phi(b)$ for some $a,b \in G$. Hence, $$cd = \phi(a)\phi(b) = \phi(ab) = \phi(ba) = \phi(b)\phi(a) = dc$$ This means $H$ is also abelian.
  • If $H$ is abelian, we have $xy = yx$, for any $x,y \in H$. Now consider two elements in $G$, say $a$ and $b$. Since $\phi$ is a mapping, we have $c=\phi(a)$ and $d=\phi(b)$ for some $c,d \in H$. Hence, $$\phi(ab) = \phi(a)\phi(b) = cd = dc = \phi(b) \phi(a) = \phi(ba)$$ Since $\phi$ is an isomorphism (more specifically injective), we have $ab = ba$. This means $G$ is also abelian.

Thanks

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    $\begingroup$ This all looks correct. Note the first part of your proof is essentially any factor group of an abelian groups is abelian, and the second part is really just saying any subgroup of an abelian group is abelian. $\endgroup$ Commented Dec 28, 2013 at 14:27
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    $\begingroup$ @PVAL What is a factor of a group? I can see why you claim the second part implies any subgroup of abelian is abelian. This is essentially because $\phi^{-1}(H)$ is a sub-group of $G$. Am I right? $\endgroup$
    – John Smith
    Commented Dec 28, 2013 at 14:32
  • $\begingroup$ en.wikipedia.org/wiki/Quotient_group You do not need to understand this to do the question, I was just mentioning what your proof gives immediately about natural constructions (of a subgroup and a factor group). $\endgroup$ Commented Dec 28, 2013 at 14:33
  • $\begingroup$ @PVAL Thanks. Let me get to Quotient groups later. I am still in the first chapter of Dummit and Foote. Quotient groups are only in the 3rd chapter. Let me take it step by step. But thanks anyway. $\endgroup$
    – John Smith
    Commented Dec 28, 2013 at 14:35
  • $\begingroup$ $\phi (G)$ is actually a subgroup of $H$. In particular $G$ is isomorphic to a subgroup of $H$. $\phi ^{-1}(H)=G$ for any map (homomorphism or not) $G \rightarrow H$ $\endgroup$ Commented Dec 28, 2013 at 14:41

1 Answer 1

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Your proof is very well written.

You've proven precisely what you needed to prove, and you did so based solely on the definitions of "isomorphism" and "commutativity," working with the tools you've acquired thus far.

Bravo!

Note:

When I first encountered "abstract algebra" and what it means for two groups to be isomorphic, I recall completing proofs to verify all the ways in which an isomorphism between groups preserves the group "structures", with commutativity being one such structural property among many.

My instructor at the time was very demanding, to the finest details, requiring justification for every assertion in a proof. He would have been pleased with your explicit appeal to the fact that, as an isomorphism, $\phi$ is surjective, and also your explicit appeal to the definition of $\phi$ as a mapping, and an injection.

Now, in the event you have such an instructor, my only suggestion would be

  • to add something like what I've included using "underbraces" below:

$$cd = \phi(a)\phi(b) = \underbrace{\phi(ab) = \phi(ba)}_{G\;\text{is abelian}} = \phi(b)\phi(a) = dc$$ $$\phi(ab) = \phi(a)\phi(b) = \underbrace{cd = dc}_{H\;\text{is abelian}} = \phi(b) \phi(a) = \phi(ba)$$

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  • $\begingroup$ @amWhy: Needs another UV +1 $\endgroup$
    – Amzoti
    Commented Dec 29, 2013 at 13:59

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