I'm studying Abstract Algebra (Group Theory) for the first time, and I found a lemma that states: Let $G$ be a group, and $a,b$ in $G$ Then $(ab)^{-1} = b^{-1}a^{-1}$
Proof: $$(b^{-1}a^{-1})ab = b^{-1}eb = b^{-1}b = e$$ $$ab(b^{-1}a^{-1}) = e \implies (ab)^{-1} = b^{-1}a^{-1}$$
My professor mentioned that the other way around is incorrect, so I tried writing out the proof, and it makes sense why it isn't correct, but I want to guarantee that I understand why it's wrong:
$$(a^{-1}b^{-1})ab = a^{-1}b^{-1}ab$$ Here, are we not allowed to cancel out the elements a and b if their inverses are not next to them? Why wouldn't it, if groups must maintain associativity?
The inverse of an element $x$ in a group is an element $y$ in the group such that $xy=e$ and $yx=e$ where $e$ is the identity. In other words, the inverse is defined in terms of a property it has with respect to $x$. The proof that $(ab)^{-1}=b^{-1}a^{-1}$ is merely checking that $b^{-1}a^{-1}$ has the correct property to be the inverse of $ab$.
When $a$ and $b$ commute, then $(ab)^{-1}=a^{-1}b^{-1}$. The problem with this in general is that one can't be certain that $(ab)(a^{-1}b^{-1})=e$ since we don't have cancellation. Say, for instance, that we wanted to cancel $a$ and $a^{-1}$ in $aba^{-1}$ so that $aba^{-1}=b$. If we multiply by $a$ on the right, we get that $ab=ba$, which is exactly the equality that shows that $a$ and $b$ commute. Therefore, we see that $aba^{-1}=b$ if and only if $a$ and $b$ commute.
If $a$ and $b$ do not commute, then $aba^{-1}=c\not=b$, and, since inverses are unique, $c^{-1}\not=b^{-1}$, so $cb^{-1}\not=e$. Therefore, $a^{-1}b^{-1}$ is not the inverse of $ab$.