Use of bijection in application of Lagrange's theorem to prove Euler's totient theorem.

Question

I'd like to gain some clarity on the following question in my textbook:

Prove Euler's theorem for finite abelian groups $G$. To do this, for any $a\epsilon G$, calculate $ \prod_{x \in G} (x) $ and $ \prod_{x \in G} \ (ax) $.

The answer is shown as follows:

For every element $a \epsilon G$ we have $ G = \{ ax \mid x\in G \} $, since $ \lambda_a : G \to G, \ x \mapsto ax $ is a bijection. Since $G$ is abelian, we have with $ n = |G|:$ $$ a^n \prod_{x\in G}(x) = \prod_{x\in G} (ax) = \prod_{x\in G} (x) $$ after simplification of $ \prod_{x\in G}(x) $, therefore $ a^n = e_G $, hence Euler's theorem for abelian groups has already been proven.

I've translated it from German, so it may sound a bit odd. It would be great if someone could provide a step-by-step more detailed explanation. There is also a similar answer to this question, but I wasn't able to make the most sense out of it either.

I would really appreciate it if someone could go into detail about the theory surrounding this answer approach, it is difficult for me to even pinpoint concrete questions at this stage, but they could be the following:

1. Where is the commutative property of the group leveraged exactly? Why would the same approach not work for a non-abelian group?
2. I understand that the bijection maps every element in one $G$ to $ax$ in a different $G$, but why does the bijection ensure that the product of the elements in the different groups need to be equal?

I also understand that there are other ways to prove the property $ a^{|G|} = e $, but I want to understand this one.

Answer 1

1. Say $G$ consists of elements $x_1,\ldots,x_n$. Then $$a^n\prod_{i=1}^nx_i = a^nx_1\cdots x_n,$$ whereas $$\prod_{i=1}^n (ax_i) = ax_1ax_2\cdots ax_n.$$ Commutativity is used to equate the two, by commuting all the copies of $a$ to the left. Also in:

2. Again, you have a bijection of elements from $G$ to itself. Because the group $G$ is abelian, multiplying the elements of $G$ will result in the same thing, regardless of the order in which they are multiplied. So even though the elements $ax_1,ax_2,\ldots,ax_n$ are the same as the elements $x_1,x_2,\ldots,x_n$, in some order (and not necessarily in an identical order), the product of all $ax_i$ is the same as the product of all $x_i$ because we have the same factors, just in a different order.