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I was wondering if someone could tell me if my method is correct. I feel like the substation $b=e$ is a stretch and that there is a better way to do it.


We know that $G$ is Abelian $\iff (ab)^{-1} = a^{-1}b^{-1}$. So I know that we will have to manipulate $aba = b$ into $(ab)^{-1} = a^{-1}b^{-1}$.

But this is where I get stuck. $$aba = b \implies ab = ba^{-1} \implies (ab)^{-1} = (ba^{-1})^{-1} \implies (ab)^{-1}=ab^{-1}$$ This is close, but not what I want. One thing that I did notice is that if $b=e$, then $aba=b \implies a^2 = e \implies a=a^{-1}$. Since $a^{-1}$ is a unique element we know that $a=a^{-1}$ regardless of what $b$ is. So if we make that substitution into out previous equation we get $$(ab)^{-1} = a^{-1}b^{-1}$$ Therefore $G$ is Abelian.

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    $\begingroup$ Notice that $ab=ba^{-1}$ and $aa=e$ for $b=a$ and just substitute the value of $a^{-1}$ $\endgroup$
    – Gio
    Jan 31, 2021 at 21:50
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    $\begingroup$ The substitution $b=e$ is correct, since $aba=b$ holds for every $a$ and $b$ in $G$. $\endgroup$
    – azif00
    Jan 31, 2021 at 21:50
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    $\begingroup$ Why is setting $b = e$ a "stretch"? It is a completely standard result that if $g^2 = e$ for all elements $g$ in a group that the group is abelian. Therefore it's really quite natural to set $b = e$ to get $a^2 = e$ for all $a$ in the group. $\endgroup$
    – KCd
    Jan 31, 2021 at 21:50

3 Answers 3

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You're correct.

If $a^2=e$ for all $a\in G$, then for any $g,h\in G$,

$$\begin{align} (gh)^2&=ghgh\\ &=e\\ &=ee\\ &=g^2h^2\\ &=gghh, \end{align}$$

from which it follows that $gh=hg$. Hence $G$ is abelian.

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The hypothesis implies an exponent of two for the group. But then it is abelian, since $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$, for any two elements of the group $a$ and $b$.

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Choose

$a = b; \tag 1$

then

$aba = b \tag 2$

becomes

$a^3 = a, \tag 3$

which implies

$a^2 = e, \tag 4$

$e$ being the identity element of $G$; thus,

$a = a^{-1}, \tag 5$

whence (2) becomes

$aba^{-1} = b, \; \forall a, b \in G, \tag 6$

or

$ab = ba, \tag 7$

and we see that $G$ is abelian. $OE\Delta$.

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