$a^{-1}b \implies ba^{-1}$ for group $H$ with $a \sim b \impliedby ab^{-1} \in H$

Question

I have an issue with the following example proof, namely the part where $(ab^{-1})^{-1} = (b^{-1})^{-1}a^{-1}$. Why is that?

I would say $(ab^{-1})^{-1} = a^{-1}b$. Nowhere does it say that the operation of $G$ is commutative, so why then would it be allowed to reverse $a^{-1}b$ to $ba^{-1}$?

Note that this is the Example 2 mentioned above:

Example 3 would make sense to me, if the operation of $G$ is addition of integers, but it does not say so specifically, which makes me unsure if it is supposed to be interpreted this way or if I am missing something.

Answer 1

I think you're mistaken about how inversion acts on products; in particular, for any group $G$ and any $a,b\in G$ we must have: $$(ab)^{-1}=b^{-1}a^{-1}.$$ If you already know that inverses are unique, this is easy to verify; if you multiply it by $ab$ (on either side), you get the identity element $e$: $$abb^{-1}a^{-1}=aea^{-1}=aa^{-1}=e$$ $$b^{-1}a^{-1}ab=beb^{-1}=bb^{-1}=e.$$ Basically, you need to reverse the order when taking an inverse in order to put the cancelling terms next to each other in the above product.

That is to say, order does matter, but inversion flips the order. The book's manipulation is therefore correct.

Answer 2

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

This is what is true in general. For it to be the case that $(ab^{-1})^{-1} = a^{-1}b$, that's when you'd need things to commute.