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An abelian group from what I know is that it's a commutative group. A group is a nonempty set with one associate binary operation that is closed, has a unity, and multiplicative inverses for each element. In that case, how would I prove the claim. What I did was to expand the left side forming $(a*b)(a*b)$ and then since it's a abelian group, I swapped the first $b$ with the second $a$ and got $(a*a)(b*b)$. Then ${a^2}{b^2}={a^2}{b^2}$. Am I correct? Also, I am new to groups, abelian groups,etc. and have been searching around for more concrete knowledge. If you can explain what groups, etc. are, that would be be greatly appreciated as well.

Shaun
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Henry Lee
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  • by commutativity $abab = (ab)^2 = a^2 b^2$ sure. – reuns May 13 '16 at 04:21
  • So I'm correct right? – Henry Lee May 13 '16 at 04:23
  • for the swapping part, sure. for the definition of a group, too. for the "what group are" part : $(\mathbb{Z},+)$, $(\mathbb{R}^*, \times)$, (any vector space, $+$), (the set of $n \times n$ invertible matrices, $\times$) all are groups. – reuns May 13 '16 at 04:24

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Your proof for if the group is abelian then $(ab)^2=a^2b^2$ is ok.

However, you are supposed to also show that if $(ab)^2=a^2b^2$ for all $a,b$ then $G$ is abelian. So assume $G$ is a group and $(ab)^2=a^2b^2$ for all $a,b\in G$, i.e., $abab=aabb$. As $G$ is a group, we hvae the inverse $a^{-1}$ of $a$. Multiplying our equation from the left with $a^{-1}$ gives $a^{-1}abab=a^{-1}aabb$, or $1bab=aabb$, or $bab=abb$. Similarly, multiply with $b^{-1}$ fromt the right to obtain $babb^{-1}=abbb^{-1}$ and finally $ba=ab$. As $a,b$ were arbitrary, $G$ is abelian.

Hagen von Eitzen
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  • Thank you for your answer and I understand it :). However, I think from what I'm reading from the paper, the question is telling us to prove the claim iff the group is abelian. I don't think the question is asking to prove the claim to show its abelian. – Henry Lee May 13 '16 at 04:59
  • @HenryLee: an if and only if statement involves proving two things: I would recommend looking up what this means, and then reread Hagen von Eitzen's answer. As it stands, you have accepted an answer which 1) gives the exact same proof as you did and 2) does not answer your question. – Alex Wertheim May 13 '16 at 05:07
  • So I have to prove both ways. Is that what you're implying? In that case, I've only proven half of it. – Henry Lee May 13 '16 at 05:40
  • @HenryLee: yes, that is what I'm saying, and indeed, you have only proven half of it. Fortunately, this nice answer proves the other half. :) – Alex Wertheim May 13 '16 at 05:59
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As for the other direction,

If $(ab)^2 = a^2b^2$ for all $a,b,$ then multiplying the equation on the left by $a^{-1}$ and on the right by $b^{-1}$ yields $ba= ab$. Implying the group is abelian.

SquirtleSquad
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let $G$ be a abelain group

$\Rightarrow ab=ba,\text{for all}\; a,b \in G$

Now,

$(ab)^2=(ab)(ab)=a(ba)b=a(ab)b=aabb=a^2b^2$

kalpeshmpopat
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