$(G,\cdot)$ is abelian $\Longleftrightarrow$ $\forall a,b \in G: (ab)^{-1}=a^{-1}b^{-1}$

Question

Show that for a group $(G,\cdot)$ the following statements are equivalent:

$A:$ $(G,\cdot)$ is abelian

$B:\forall a,b \in G: (ab)^{-1}=a^{-1}b^{-1}$

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$A \Longrightarrow B:$

Since $(G,\cdot)$ is a group:

$\forall a,b \in G: (ab)^{-1}=b^{-1}a^{-1}$

and since $(G,\cdot)$ is commutative:

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

$B \Longrightarrow A:$

Since $B$ holds:

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

Suppose $(G,\cdot)$ is not abelian, then:

$aba^{-1}b^{-1}$ has to equal $e$ which would mean, that since $(G,\cdot)$ is associative,

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

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

And since $(aba^{-1})b^{-1}=e \Longrightarrow (aba^{-1})=b$

Which means $ba^{-1}=a^{-1}=a^{-1}b^{-1}$ and $ab=b=ba^{-1}$

in this case we can calculate $(ab)(a^{-1}b^{-1})=ba^{-1}$ which supposed to be $e$.

But since in a group the inverse has to be specific and unique, this is a contradicition!

Since weve choosen $aa^{-1}=e\,\,\,\wedge bb^{-1}=e$

$\Longrightarrow (G,\cdot)$ has to abelian

$B\Longrightarrow A$

So concluding to $A \Longleftrightarrow B$

$\Box$

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Could someone look over it? Last exercise on group theory I messed up as some of you have seen already :) Is this correct this time?

Answer 1

Your $A\implies B$ is fine. I would write the last line as $$ (ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1} $$ because that better follows the flow of your argument. But that's a minor aesthetic concern.

For $B\implies A$, I think you need to put more emphasis on the $\forall$. Also, you go from $(ba^{-1}b^{-1})=a^{-1}$ to $ba^{-1}=a^{-1}$, which is just not true.

I would personally choose to go contrapositive, as a first attempt. That is, show $\lnot A\implies \lnot B$. So assume the group isn't abelian, and find some $a,b$ which do not fulfill $B$. So that's what I'll do.

$\lnot A\implies \lnot B$: Since the group is non-abelian, by definition there is some pair of elements $a,b$ such that $ab\neq ba$. Multiply both sides from the right by $a^{-1}b^{-1}$, and you get $aba^{-1}b^{-1}\neq e$, and we're done.

But if you want to prove it directly, that's very doable too: we have, for each $a,b\in G$, that $$ aba^{-1}b^{-1}=e\\ aba^{-1}=b\\ ab=ba $$ proving that the group is abelian.

This might shed some light on why $aba^{-1}b^{-1}$ is called the commutator of $a$ and $b$: it measures whether $a$ and $b$ commutes.

Answer 2

An easier proof of the second part will be will be : $ab$ = $(a^{-1})^{-1}.(b^{-1})^{-1}$ = $(a^{-1}b^{-1})^{-1}$ (why ?) = $((ba)^{-1})^{-1}$ (why? ) = $ba$

So the group is abelian. Just joined the community, yet to learn latex properly.