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I had proved the statement as the following,

Suppose $a,b \in G$ and $a^{-1} = a$ thus $b^{-1} = b$.

Consider $a^{-1}b^{-1}$.

Let $e$ be an identity element in $G$ this will get $baa^{-1}b^{-1} = beb^{-1} = bb^{-1} = e$ .

Thus $a^{-1}b^{-1} = (\, ba )\, ^{-1}$

Since $(\, ba )\, ^{-1} = ba$, $ab = a^{-1}b^{-1} = (\, ba )\, ^{-1} = ba$

$\therefore G$ is an abelian group. Quod Erat Demonstrandum

Can you check the errors for me please?

Supakorn Srisawat
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    The reasoning is okay. Note that $(ba)^{-1}=a^{-1}b^{-1}$ holds in any group, so you don't need to show this separately. If you do want to show that, you need to show that $baa^{-1}b^{-1}=e$ and $a^{-1}b^{-1}ba=e$. You forgot one side in the above argument.

    Note that the assumption $x=x^{-1}$ for all $x\in G$ yields that $ba=(ba)^{-1}$. This is the fundamental observation that you need in order to prove this statement.

     – Mathematician 42 Aug 04 '20 at 07:36
  • @Mathematician42 you don't need to show that the inverse is two-sided as you are already in a group, and inverses are always two-sided. – David A. Craven Aug 04 '20 at 09:07
  • @Chrystomath the assumption is that $x=x^{-1}$ for all $x$, and hence in particular $(ba)^{-1}=ba$. No proof required, surely? – David A. Craven Aug 04 '20 at 09:08
  • @DavidA.Craven: I stated that $(ba)^{-1}=a^{-1}b^{-1}$ holds in any group. If you want to show it, you need to show that $a^{-1}b^{-1}$ is both the left and right inverse to $ba$. You need to show a two-sided statement by definition, whether it holds automatically for a different reason in a group does not devalue the point I made. By that logic we can discard the statement $(ba)^{-1}=a^{-1}b^{-1}$ altogether as it holds in any group and you don't need to show it. Your statement is itself a good exercise for the poster but I doubt it's in his standard toolkit at this point. – Mathematician 42 Aug 04 '20 at 10:27
  • @Mathematician42 surely that one-sided inverses are two-sided is the third statement you prove in a group theory course, well before orders of elements, abelian, anything. Statement 1 is that the identity is unique, statement 2 is that inverses are unique, and then you prove that groups are cancellative. Do you think otherwise? (I admit I've never taught a first course in group theory.) – David A. Craven Aug 04 '20 at 10:36
  • @DavidA.Craven: I don't think otherwise and I don't think we are clashing over anything here :) I'm just saying that if he felt the need to 'prove' $(ba)^{-1}=a^{-1}b^{-1}$, then assuming that one-sided inverses are two-sided inverses (in a group) is obvious to him, might be a stretch. I'd say both statements are equally obvious if you're used to working with groups. The OP is not yet familiar with groups and is still learning, let's not assume such statements are obvious to him at this point. – Mathematician 42 Aug 04 '20 at 12:23

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It appears to be good to me. I would only make one edit by beginning the proof with

Suppose for all $a \in G$, $a = a^{-1}$.

Now suppose we have $a, b \in G$. Then $a = a^{-1}$ and $b = b^{-1}$.

Doctor Who
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