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Can I say this:

Consider $g h h g = 1_G$

We know that $g^2 = 1_G$ and so we get

$ g h h g = g h^2 g = g g = g^2 = 1_G $

As every group must have a unique inverse, in order for my "consider" claim to hold, we must have that $gh = hg$ and therefore the group is abelian.

Is this correct?

EDIT: What's wrong with my proof then? I've seen the $(ab)^2 = 1_G$ proof and I understand that but why is that correct and my one wrong? What am I missing out on proving?

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    It is right. But I guess it is better to write $1_G$.2012-12-28

5 Answers 5

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Your idea works, but your wording of it is somewhat nonstandard. The initial "consider" is confusing, because that word is usually followed by a definition for-the-purpose-of-the-proof -- but you're not really defining anything there.

It would be clearer if you just write

For any $g$ and $h$ we have $ghhg=gh^2g=g1g=g^2=1$. On the other hand, $ghgh=(gh)^2=1$. Combining these we get $ghhg=ghgh$, and canceling $gh$ on the left, we get $hg=gh$. Since $g$ and $h$ were arbitrary, this proves that $G$ is abelian.

There is no need to speak of inverses or their uniqueness explicitly.

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Hints:

$\forall\,x\in G\,\,,\,x^{-1}=x$

$\forall\,x,y\in G\;\;,\;\;1=(xy)^2=xyxy\ldots$

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    MSE occasionally turns into YouTube comment section in its silly pettiness.2012-12-28
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$abab=1_G$ $ababba=1_Gba$ Since $abba=aa=1_G$ $ab=1_Gba=ba$

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To show $ab=ba,\forall a,b\in G$it suffice to show $1=(ab)(ba)$ since $ab=(ab)^{-1}$.

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The condition $g^2 = e$ is equivalent to $g = g^{-1}$ for all $g\in G$.

Now assuming this condition we have that for any $a,b\in G$:

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

Hence $G$ is abelian.