MHB Is a Group Abelian if Inverses Commute?

[cbarker1]

Dear Everyone,

Here is the problem that I am attempting to prove:

"Prove that a group $(G,\star)$ is abelian if and only if ${(a\star b)}^{-1}={a}^{-1}\star {b}^{-1}$ for all a and b in $G$."

My attempt:
Let $(G,\star)$ be a group $G$ under the binary operation $\star$. Then suppose $G$ is abelian. Then we know that $G$ has the associative property; there is an identity element $e$ in $G$; there is an inverse element for each $x$ in $G$. We know that $\star$ is commutative under $G$. So for all $a,b\in G$, $(a\star b)\in G$. Then ${(a\star b)}^{-1}\in G$. Then, by proposition 1 ( abstract Algebra, Dummit Foote second edition page 18), ${(a\star b)}^{-1}={b}^{-1} \star {a}^{-1}$. But we that $G$ is a abelian, so ${b}^{-1} \star {a}^{-1}={a}^{-1} \star {b}^{-1}$.

I am having trouble with the converse of this problem. Any suggestions on my attempt will be appreciated.

Thanks,
Cbarker1

 

[Ackbach]

So, assume $(ab)^{-1}=a^{-1}b^{-1}$ for all $a,b\in G$. We want to show that $ab=ba$ for all $a,b\in G$. We have that
$ab(ab)^{-1}=e,$ but it's also the case that $abb^{-1}a^{-1}=e$, by the same theorem you quoted earlier. By assumption, $aba^{-1}b^{-1}=e$. What we want to show is equivalent to $b^{-1}aba^{-1}=e$. Take this:
\begin{align*}
ab&=ab \\
b^{-1}aba^{-1}&=b^{-1}aba^{-1} \\
b^{-1}aba^{-1}&=(a^{-1}b)^{-1}ba^{-1} \\
b^{-1}aba^{-1}&=ab^{-1}ba^{-1} \\
b^{-1}aba^{-1}&=aa^{-1} \\
b^{-1}aba^{-1}&=e,
\end{align*}
as required.

 

[Olinguito]

By Dummit & Foote Proposition 1, $(ba)^{-1}=a^{-1}b^{-1}$. If also $(ab)^{-1}=a^{-1}b^{-1}$, then
$$\begin{align*}(ba)^{-1} &= a^{-1}b^{-1} = (ab)^{-1} \\\\ ab(ba)^{-1} &= ab(ab)^{-1} = e \\\\ [ab(ba)^{-1}]ba &= eba = ba \\\\ ab[(ba)^{-1}ba] &= ba \\\\ ab &= ba\end{align*}$$
showing that the group is Abelian.