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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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    Do you mean $1_G$ when you write $1_g$? If not, what does $1_g$ mean?2012-12-28
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    Yeah, sorry, its supposed to be the identity of group G. Typed it too quickly2012-12-28
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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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    Well, I seem to have today a serial downvoter on my tail, stacking me...I wonder why.2012-12-28
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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.