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Let $G$ be a group. Show that if every non-identity element in $G$ has order $2$ then $G$ is abelian.

Proof:

Let $a,b $ be non-identity elements in $G$. Since $|a|=|b|=2$ , that means $ab=babaab$ $=$ $ba$.

Is the proof correct? How can I improve it?

TheSimpliFire
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    The proof is correct as written. – rubikscube09 May 04 '19 at 21:04
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    I must be tired by I don't see how to justfify the first equality. – elidiot May 04 '19 at 21:08
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    @elidiot (baba)ab=(e)ab=ab. bab(aa)b=babb=ba –  May 04 '19 at 21:09
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    It uses $|ba|=2$ on the left. – Berci May 04 '19 at 21:10
  • I flagged as a duplicate both because the problem in question is duplicated and because the suggested solution is duplicated (it is the same as Gudesa Kuse's at the linked dupe, so the question of "is this right" is also answered at that question.) – rschwieb May 06 '19 at 13:47
  • @rschwieb what I wrote is my own work. If this is similar to any other post, then that is merely a coincidence. The question might have been asked, but I have tagged the proof verification to verify the proof that I have written. Lastly, the proof, although seemingly trivial, has been posted to reassure myself that the logic behind the implications is correct. Lastly, it is not beneficial to look at other solutions prior to writing one. Therefore, I presume that there are subtle differences (if this work is similar). –  May 07 '19 at 13:56
  • @topologicalmagician Hi, I know it's your own work, and I know you didn't want to look before writing your solution. But what I am proposing is that you look for duplicates after you've written your solution and before you post it as a question. The idea is that if you find a duplicate and your solution is new, you'd post it at the duplicate as a solution. If not new, you would have already answered your own question, eliminating the need to duplicate it. I don't see any conflict with this and the desires you expressed. – rschwieb May 07 '19 at 14:02
  • @topologicalmagician Please also take a look at this meta discussion from this past year for some context on why I'm suggesting this. And if your question appeared nowhere on the site, then I'd suggest posting it as a "here's a problem I'm solving, and I'm putting my solution below" format. That's also mentioned at the meta post. We're trying to minimize questions of the form "is this right?" – rschwieb May 07 '19 at 14:05
  • @rschwieb Thank you very much, I understand. Out of curiosity, How do you propose I get feedback if I don't post it? –  May 07 '19 at 14:06
  • @topologicalmagician Believe me, if you post an incorrect solution, people will be more than happy to point out the problems. That's the point, right? – rschwieb May 07 '19 at 14:07
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    @rschwieb thanks, i'll keep that in mind next time. –  May 07 '19 at 14:07
  • @topologicalmagician Thanks for considering what I've said. I do understand your intent and position, and I'm sorry if I sounded unfriendly in any way. It's just that after seeing the big picture for a while with lots of these questions, one starts to see their problematic nature, and how they might be better handled. The alternative way is perhaps not the most obvious :) – rschwieb May 07 '19 at 14:09
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    @rschwieb one last question, as I am relatively new to this site, if I post my solution as a duplicate at another post, are the chances of receiving feedback, as high as when making a new post? –  May 07 '19 at 14:17
  • Good question @topologicalmagician ! New solutions to old problems bump the old question to the top of the Activity list, granting it renewed visibility. And I also anecdotally feel that people are more critical towards new answers to old questions ("could this person really have added something new that the other answers didn't have?" or "wow look, they tried to answer this question nobody attempted before for a long time") . So I think the chances are good for a new answer on an old question. – rschwieb May 07 '19 at 14:38

2 Answers2

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Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^{-1}=ab$, but $(ab)^{-1}=b^{-1}a^{-1}=ba$ by using twice again the remark.

elidiot
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Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).

Shaun
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