Can someone explain this proof about groups? (Let $G$ be a group. If $a, b \in G$, then $(ab)^{-1} = b^{-1}a^{-1}$.) I just started to read this file, and I am confused about the  proof of this proposition.

Proposition 2.4. Let $G$ be a group. If $a, b \in G$, then $(ab)^{-1} = b^{-1}a^{-1}$.
Proof. Let $a,b \in G$. Then $abb^{-1}a^{-1} = aea^{-1} = aa^{-1} = e$. Similarly, $b^{-1}a^{-1}ab = e$. But by the previous proposition, inverses are unique; hence, $(ab)^{-1} = b^{-1}a^{-1}.\qquad\square$

To clear what "previous  proposition" is,

Proposition 2.3. If $g$ is any element in  a group $G$, then the inverse of $g$, $g^{-1}$, is unique.

Also,

Most of the time we will write $ab$ instead of $a \circ b$; however, if the group already has a natural operation such as addition in the integers, we will use that operation.


I am confused about $abb^{-1}a^{-1}$ and $b^{-1}a^{-1}ab$. Is it assumed that $G$ is abelian? Also, what is the "order of operation": is it left to right or right to left?
Note: A similar question can be found, but this is not a duplicate. I am asking about the commutativity of the group, not how is the proposition proven.
 A: It is not necessary that $G$ be abelian, this follows from the previous proposition and group axioms.
Recall that for any group $G$ with a group operation $\cdot,$ the group operation must be associative, so for any $a,b,c \in G, (a \cdot b) \cdot c = a \cdot (b \cdot c).$ So although each operation must be taken left to right (the operation is only commutative if $G$ is abelian) the order that the operations are taken in should not matter.
So, when computing $abb^{-1}a^{-1}$ we can rewrite this as $a(bb^{-1})a^{-1},$ which is the same as $a(e)a^{-1}$ because by definition $bb^{-1} = e.$ This can then again be rewritten as $(ae)a^{-1} = aa^{-1}$ by the definition of the identity element, and then this is just $e,$ again by the definition of the inverse.
A: No, $G$ is not assumed to be abelian.
Recall that uniqueness of inverses states that for each $g\in G$, whenever $h\in G$ such that $gh=e=hg$, then $h=g^{-1}$.
It is shown in the proof in question that $(\color{red}{ab})(\color{blue}{b^{-1}a^{-1}})=e=(\color{blue}{b^{-1}a^{-1}})(\color{red}{ab})$. But $ab\in G$. Hence $$(\color{red}{ab})^{-1}=\color{blue}{b^{-1}a^{-1}}.$$
A: It is absolutely not assumed the group is abelian. It is because the group might not be abelian that this is not a trivial question.
But addition is ASSOCIATIVE.
So $a*b*c*d$ can be don in any of the following orders.
$(((ab)c)d$ or $(ab)(cd)$ or $(a(bc))d$ or $a((bc)d)$ or $a(b(cd))$.  (I think that is all[1]).  You can group them in pairs any way you want but you must pair them only with terms consecutively and you can not switch the order within any pairs.
So using $(ab)(cd) = (a(bc))d= a((bc)d)$
$(ab)(b^{-1}a^{-1}) = (a(bb^{-1})) a^{-1}= a((bb^{-1})a^{-1})$ and that's that then.... it follows
$(ae)a^{-1} = a(ea^{-1})=$
$aa^{-1} = aa^{-1} = e$.
Thus the solution to $(ab)* x = e$ is $x = b^{-1} a^{-1}$ and the previous proposition says this inverse is unique.
So $(ab)^{-1} = b^{-1}a^{-1}$.
...........
Now if we assumed the group WERE abelian the answer is trivial:  $(ab)x = e$ we need to get rid of the $a$ so we need a $a^{-1}$ and we need to get rid of the $b$ we need a $b^{-1}$ and as the group is abelian order doesn't matter so just do $a^{-1}b^{-1}$.
If we have $(ab)(a^{-1}b^{-1})$ just group the $a$ with $a^{-1}$ and the $b$ with $b^{-1}$ to get $aa^{-1}bb^{-1}= ee =e$.
But because the group needn't be abelian the "just group them together" argument is not valid.
However placing the inverses butt to head right to left to counter left to right allows us to unpeel like a grape.
Pairing $abcdfgh$ with $h^{-1}g^{-1}f^{-1}d^{-1}c^{-1}b^{-1}a^{-1}$ will give us
$(abcdfgh)(h^{-1}g^{-1}f^{-1}d^{-1}c^{-1}b^{-1}a^{-1}) =$
$a(b(c(d(f(g(hh^{-1})g^{-1})f^{-1})d^{-1})c^{-1})b^{-1})a^{-1}=$
$a(b(c(d(f(gg^{-1})f^{-1})d^{-1})c^{-1})b^{-1})a^{-1}=$
$a(b(c(d(ff^{-1})d^{-1})c^{-1})b^{-1})a^{-1}=$
$a(b(c(dd^{-1})c^{-1})b^{-1})a^{-1}=$
$a(b(cc^{-1})b^{-1})a^{-1}=$
$a(bb^{-1})a^{-1}=$
$aa^{-1}=e$
======
Post script
[1] More detail.
Suppose we have $a*b*c*d$
And $a*b = \alpha$
$b*c = \beta$
$c*d = \gamma$
$\alpha*c = H$
$a*\beta = H$ (Note:  Associativity requires that $a*\beta =a*(b*c)$ and $\alpha*c = (a*b)*c$ both equal the same value $H$)
$\beta* d=K$
$b*\gamma = K$ (Note: ditto.  $\beta*d=(b*c)*d$ must be equal to $b*(c*d) =b*\gamma$)
And $\alpha*\gamma =a*K=H*d=N$.
Then we can interpret $a*b*c*d$ in any of the following six ways. (Note: All of them are "left to right" the only difference is how we group the pairs.)  (Also note: 2) and 5) are both $(a*b)(c*d)$ so they can be consider the same if we don't worry about whether we spell out $a*b= \alpha$ or $c*d = \gamma$ first.)

*

*$a*b*c*d=a*b*(c*d) = a*b*\gamma = a*(b*\gamma) =a*K =N$.


*$a*b*c*d=a*b*(c*d) = a*b*\gamma = (a*b)*\gamma =\alpha * \gamma =N$.


*$a*b*c*d=a*(b*c)*d = a*\beta*d = a*(\beta*d) = a*K = N$.


*$a*b*c*d=a*(b*c)*d = a*\beta*d = (a*\beta)*d = H*d = N$.


*$a*b*c*d = (a*b)*c*d = \alpha*c*d = \alpha*(c*d)= \alpha*\gamma =N$


*$a*b*c*d = (a*b)*c*d = \alpha*c*d = (\alpha*c)*d= H*d =N$
