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I didn't see the similar question. My question isn't a duplicate because I understand how to prove this. But I don't understand where the hint comes from. That question doesn't flesh it out.

  1. Pretend the question didn't give the hint. How can you prognosticate (please see my profile) to Consider $(a*b)*(a*b)$? This trick is too magical and superhuman. I didn't see it.

  2. The other trick is to write $e *e = (a*a)*(b*b)$. How can you prognosticate this?

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    $\begingroup$ I have to admit, my first thought on seeing the title was "Prove every involutive group abelian with this one weird trick!" $\endgroup$ – Steven Stadnicki Dec 23 '13 at 17:25
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    $\begingroup$ @Steven: The weird old tip that math professors don't want you to know! $\endgroup$ – Henning Makholm Dec 23 '13 at 17:30
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    $\begingroup$ You prognosticate it carefully. $\endgroup$ – Alexander Gruber Dec 23 '13 at 17:32
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    $\begingroup$ Also ( merriam-webster.com/dictionary/prognosticate ) ( merriam-webster.com/dictionary/prognosis ) You keep using that word. I do not think it means what you think it means. [Note syn: prophesy] $\endgroup$ – Eric Stucky Dec 24 '13 at 1:36
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    $\begingroup$ This is not a duplicate of said question because it specifically asks not for a proof (which the OP knows) but, instead, for some intuition behind successful strategies for constructing such proofs - something that is not addressed at all in the proposed duplicate. Probably there is a good chance of further helpful answers appearing if it is reopened. $\endgroup$ – Bill Dubuque Dec 24 '13 at 3:37
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A key idea is clearer this way: $\ x\cdot x = e\,\Rightarrow\, x \color{#c00}{\overset{(1)}=} x^{-1}\, $ so $\ ab \color{#c00}{\overset{(1)}=} \color{blue}{(ab)^{-1}}\! \color{#0a0}{\overset{(2)}=} b^{-1} a^{-1}\! \color{#c00}{\overset{(1)}=} b a\ \ \,$ QED

Now it is clearer where the proof "comes from", namely it arises by "overlapping" the identities $x\! \color{#c00}{\overset{(1)}=} x^{-1}$ and $\,(xy)^{-1}\color{#0a0}{\overset{(2)}=} y^{-1} x^{-1}$ i.e. by discovering some term $\,\color{blue}{(ab)^{-1}}\!$ where both identities apply. Then, just as we did above, we can rewrite the term in two ways, giving a possibly new equality.

This is a widely applicable method of deriving consequences of identities, i.e. by "unifying" terms of both so that both identities apply. In fact in some cases it can be used to algorithmically derive all of the consequences, so yielding algorithms for deciding equality, e.g. see the Knuth-Bendix equational completion algorithm and the Grobner basis algorithm, and see George Bergman's classic paper The Diamond Lemma in Ring Theory.

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    $\begingroup$ Thanks a lot. I upvoted. I really like your colors which help me to see the different equations and parts. Please keep up your great answers! And thank you very much for your supporting comment on how this is not a duplicate. $\endgroup$ – Group Theory Dec 24 '13 at 8:55
  • $\begingroup$ See here for another example: uniqueness of identity (neutral) elements. $\endgroup$ – Bill Dubuque Aug 8 '14 at 20:04
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Let $G$ be a group, to show it is abelian is the same as to show that for $a,b\in G$, one has $ab=ba$, which is the same as $aba^{-1}b^{-1}=e$.

Now in this specific group, $a^{-1}=a$ and $b^{-1}=b$, so the last equation becomes $(ab)(ab)=e$. That is why we consider $(a*b)*(a*b)$.

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We are just given that $x^2=e$ for all $x\in G$. I am not a superhuman and right now I am guessing and wanting just to manipulate the elements. You want to see $ab=ba$ for all $a,b\in G$. Let's multiply both sides from the right(or left) by $ab$ for example. We get: $$(ab)*(ab)=(ab)*(ba)$$ Since $(ab)^2=e$, so $$(ab)*(ba)=e\to a*b^2*a=e\to e=e$$ This is a good point to start and we understand why the first key was given to us. That's it.

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To break down BS's answer even more: I want to show that $ab=ba$ for every two elements $a$, $b$. This identity looks kind of awkward, so I'm going to peel terms off of one side. First, let me multiply on the left by $a^{-1}$ (or, since I know that $a^{-1}=a$, by $a$): $aab=aba$, or $b=aba$. Progress! But I'm still not done, so let's multiply both sides on the left by $b$. Well, this gives me $bb=baba$, or $e=baba$. But this last is $e=(ba)(ba)$, and I can see that that's an identity because of the group's definition; this must mean that my original statement, $ab=ba$, was also an identity, and so the group is abelian.

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  • $\begingroup$ Thanks. I upvoted. What do you mean "my original statement, $ab=ba$, was also an identity"? Do you mean $ab = ba = e$, but where do you get this from? $\endgroup$ – Group Theory Dec 24 '13 at 8:54
  • $\begingroup$ And because you started from the answer $ab = ba$, does your work actually show the question is actually an iff statement? $\endgroup$ – Group Theory Dec 24 '13 at 8:56
  • $\begingroup$ @FrankMuer I don't mean $ab=ba=e$; I mean just $ab=ba$, the thing I want to prove. Because I 'started with the answer', got to an identity, and all my steps are reversible, I can do exactly that - reverse them - and use it as a proof of the statement. $\endgroup$ – Steven Stadnicki Dec 24 '13 at 14:37

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