Abelian groups: proving $\prod\limits_{g\in G}g=\prod\limits_{\substack{g\in G\\g^2=1}}g$

For any two different elements $a$, $b$ such that $ab=1$, pair off $a$ and $b$. So you have a bunch of couples, and some loners. The product of all elements that have been coupled off is $1$. What's left is the people who are their own inverses. Note that $g$ is its own inverse iff $g^2=1$.

We have $$\prod_{g\in G}g = \prod_{\substack{g\in G\\g^2=1}}g \cdot\prod_{\substack{g\in G\\g^2\neq1}}g $$ In the product $\displaystyle\prod_{\substack{g\in G\\g^2\neq1}}g$ every element cancels with its inverse.

$\prod\limits_{g\in G}g=\prod\limits_{g\in G}_{g^2=1}g = g_1 g_2 \dots g_n$

For some elements in the product $\prod\limits_{g_i \in G}g_i $ we have $g_i g_j = 1$. Since the group is Abelian you can arrange the product so that $g_i$ and $g_j$ are "next to each other" in the product. Then

$ g_1 g_2 \dots g_n = g_{i_1} g_{j_1} g_{i_2}g_{j_2} \dots g_k \dots g_{k+k'} \stackrel{\ast}{=} g_k g_{k+1} \dots g_{k +k'} = \prod\limits_{g\in G}_{g^2=1}g$.

where in $\stackrel{\ast}{=}$ we use that $g_{i_1} g_{j_1} = 1 = g_{i_l}g_{j_l} $ for all elements where $g_i \neq g_i^{-1}$.