Let $(G, \circ)$ be a finite abelian group with neutral element $e$.
$1.$ Show that in general $\prod_{g \in G}g^n\neq e$ for $n=1$
My idea:
I used $\mathbb Z/2\mathbb Z$, it follows that $\prod_{g \in G}g=[0]+[1]\neq[0] \in \mathbb Z/2\mathbb Z$
$2.$ Show that there exists an $n \in \mathbb N$ so that $\prod_{g \in G}g^n=e$
I found a solution but I do not understand it. It states
$\prod_{g \in G}g^2=\prod_{g \in G}(g\cdot g^{-1})$
I do not understand this step... Why is $g^{2}=g\cdot g^{-1}$?
Thank for your help
It is not true that $g^2 = g \circ g^{-1}$. Rather, each element $g$ appears in the product twice, so we let every element "appear once as itself" and "appear once as an inverse, namely its inverse's inverse". For example we can write \begin{align} \prod_{g \in G} g^2 &= \prod_{g \in G}g \circ (g^{-1})^{-1}\\ &= \prod_{g \in G} g \times \prod_{g \in G} (g^{-1})^{-1} \\ &= \prod_{g \in G} g \times \prod_{h \in G} h^{-1} \\ &= \prod_{g \in G} g \times \prod_{g \in G} g^{-1} \\ &= \prod_{g \in G}g \circ g^{-1}. \end{align}