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$G$ is a group with the property that the square of every element is the identity, then $G$ is abelian. Is my proof correct?

Attempt:

For every $a \in G, \space a^2 = e$ where $e$ is the identity element of $G$.

$a^2 = e \implies a = a^{-1}$**

$G$ is abelian if $\forall a,b \in G, \space ab=ba$

So we proceed:

if $ab = ba$

$ab\times a^{-1}b^{-1}=e$

By **

$\implies ab\times ab = e$ $\implies (ab)^2 = e$

By hypothesis this is true.

~ My question is; is this correct and also why does the $a$ from the right side get moved over first? Would I be able to write

if $ab = ba$

$ab\times a^{-1}b^{-1}=e$

as

if $ab = ba$

$ab\times b^{-1}a^{-1}=e$

with the $b$ first or does the order of moving elements over matter?

Mathstatsstudent
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  • If "$G$ is abelian" is what you want to prove, why do you beg the question and say "if $ab=ba$? – J. W. Tanner Sep 22 '19 at 21:45
  • No.... that isn't correct. You are attempting to prove that if the particular $a$ and $b$ that you chose commute it is compatible with this group property. You need to show that this group property forces that any $a$ and $b$ will commute. – fleablood Sep 22 '19 at 22:05

3 Answers3

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Note that every element is its own inverse.

Therefore,

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

That is the group is Abelian.

Mohammad Riazi-Kermani
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We know $(ab)(ab) = abab =e$. Consider the product $w=ababba$. On the one hand, $w=(abab)(ba) = e(ba)=ba$. On the other, $w=ababba = aba(bb)a=abaea=abaa = ab(e)=ab$.

kimchi lover
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Simple non-general case.

$e = (ab)(ab) = bb =aa$.

So

$ab = a*e*b = a*[(ab)(ab)]*b=$

$(a*a)(ba)(b*b) = e*(ba)*e = ba$.

Ta-da!

....

More general but probably more usefull is

For any $k \in G$ then $k^2 = k*k =e$. So by definition $k^{-1} = k$ for every $k$.

So $(ab)^{-1} = ab$.

Now it's a basic identity of every group that

$(ab)*(b^{-1} a^{-1}) = a*(b*b^{-1})*a^{-1}= a*e*a^{-1} = a*a^{-1} =e$.

So $(ab)^{-1} = b^{-1} a^{-1}$.

So in this group we have:

$ab = (ab)^{-1} = b^{-1} a^{-1} = ba$.

fleablood
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