Let $G$ be a group and let $a,b$ be in $G$. Show that $(a*b)'=a'*b'$ if and only if $a*b=b*a$.
So far, it has been proved in the text I am using that the inverse element in a group is unique and a corollary that for a group $G$, for all $a,b$ in $G$, it follows that $(a*b)'=b'*a'$.
So my attempt at a proof:
$(\implies)$ If $(a*b)'=a'* b'$ then we also have $(a*b)'=b'*a'$ (by the corollary), so $$(a*b)'=a'*b'=b'*a'=(b*a)'.$$ Since the inverse element in a group is unique, $a*b=b*a$.
$(\,\Longleftarrow\,)$ If $a*b=b*a$, then $$(a*b)'*(a*b)=(a*b)'*(b*a)=e.$$ By the uniqueness of the inverse element in a group, we must have that $(a*b)'=(b*a)'$, hence, $(a*b)'=a'*b'$.
Is this a valid proof? Can we reason by the uniqueness of the inverse element both ways?