Proof for this identity in an abelian group I need to show the following identity in an abelian group.
$$\Bigl(\prod_{g \in G}g\Bigr)^2 = e$$
I think I have the basic idea for the proof, in that you can reorder the factors of the product such that each element is besides it's inverse and then all of them cancel out. But I don't really know how I would go about making an actual proof out of that.
 A: Your idea is solid.  To write it out, I'd let $h_i=g_i^{-1}$ for each $i$.  Then write $$\left( \prod_{i=1}^ng_i\right)^2=\prod_{i=1}^ng_i\times \prod_{i=1}^nh_i=\prod_{i=1}^ng_ih_i=\prod_{i=1}^n e=e$$  Where we have used both the associative and commutative properties of group multiplication. Here $n$ is the order of $G$...note that infinite products are not even defined for a general infinite group.
And we are done.
Note:  if you are unsure of the second equality (surely that's the only one that isn't automatic, yes?) then prove it by induction on $n$.  Trivially true for $n=1$ and for higher $n$ we can divide both left and right by $g_nh_n$ to get the inductively known statement for $n-1$.  Again, we need commutativity to get what we need.
A: Your idea seems fine, provided that your group is finite. This (as in most proofs) depends on how formal you want to be. An abelian group is commutative. This property is the base case for the following inductive statement on $n \in \mathbb{N}$: given a permuation $\sigma \in \mathbb{S}_n$, we have that
$$
g_1 \cdots g_n = g_{\sigma(1)} \cdots g_{\sigma(n)}.
$$
This is just a fancy way of saying that we get for free to reorder any amount of (finite) terms, from the commutativity of the group. The inductive step is to use the proposition on $n-1$ for the first $n-1$ factors and afterwards to the latter $n-1$. It could also be skipped altogheter since it is a rutine verification. Finally, your product of $2|G|$ elements can be listed and reordered so that inverses are adjacent (this, again, could be formalized if desired). To conclude, use associativity to cancel each pair of factors.
