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I tried to solve this by following:

  1. Since $G$ is a group, an inverse exists for every element in $G$.
  2. Multiply by inverse to $a$ on both sides of $a^2 = a$.

  3. We will get $a = i$, where $i$ is the identity element.

  4. This holds true for all $a*$, which implies $G$ contains only one distinct element i.e. $i.$
  5. Hence $G$ is abelian.

Is my approach correct?

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    $\begingroup$ you are correct $\endgroup$
    – mm8511
    Aug 22, 2017 at 3:39
  • 2
    $\begingroup$ Correct, but it seems too easy for an exercise. Are you sure the question didn't say, let $G$ be a group such that $a^2=i$ (the identity element) for all $a\in G,$ is $G$ an Abelian group? $\endgroup$
    – bof
    Aug 22, 2017 at 3:42
  • $\begingroup$ I solved for a^2 = i. This I could solve easily so I got confused that there might be some subtle point I am missing. Thank you $\endgroup$
    – sourav
    Aug 22, 2017 at 3:46
  • $\begingroup$ If the goal of your question is mainly to ask about correctness of your proof (as opposed to asking for any proof of this fact), you should add (proof-verification) tag to make this clear. $\endgroup$ Aug 22, 2017 at 5:30
  • $\begingroup$ Related: Trick to proving a group has exactly one idempotent element - Fraleigh p. 48 4.31. $\endgroup$ Aug 22, 2017 at 5:33

1 Answer 1

2
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Yes, I think you are right.

Also, we can get the following.

$$abab=ab$$ gives $$aba=a,$$ which gives $$ab=ba=e.$$ Done!

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    $\begingroup$ @1ENİGMA1 Yes of course. There are very many similar easy problems. $\endgroup$ Aug 22, 2017 at 7:31

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