If $G$ is an Abelian group with odd order, then the product of all its elements is the identity. I know this problem has been asked a couple times in the site, but this is a proof verification question.
Attempt:
Let $G$ be an Abelian group of odd order. We know that $\forall g\in G$, $\vert g\vert=n$ divides $\vert G\vert$. Thus $n$ cannot be even. Since $\forall g\in G$, $\langle g\rangle \leq G$ is a cyclic group; $\langle g\rangle=\{e,g,g^2,\cdots,g^{n-1}\}$ observe that
\begin{align}
egg^2\cdots g^{n-2}g^{n-1}=(gg^{n-1})(g^2g^{n-2})\cdots (g^{\frac{n-1}{2}}g^{\frac{n+1}{2}})=e\quad (*)
\end{align}
Note that $n$ not being even implies $\frac{n-1}{2},\frac{n+1}{2}\in\Bbb{N}$. Since $G=\bigcup_{i=1}^{\vert G\vert}\langle g_i \rangle$, the product of all elements in $G$ is equivalent to the form $e(g_{1}g_{1}^{n-1}\cdots g_{1}^{\frac{n-1}{2}}g_{1}^{\frac{n+1}{2}})\cdots (g_{z}g_{z}^{n-1}\cdots g_{z}^{\frac{n-1}{2}}g_{z}^{\frac{n+1}{2}})$ where $z=\vert G\vert$. Then as shown in $(*)$ this is equal to the identity.
 A: The idea behind your proof is correct, but there are some wee peccadilloes in the mix. When you write out the long product
$$
e(g_{1}g_{1}^{n-1}\cdots g_{1}^{\frac{n-1}{2}}g_{1}^{\frac{n+1}{2}})\cdots (g_{z}g_{z}^{n-1}\cdots g_{z}^{\frac{n-1}{2}}g_{z}^{\frac{n+1}{2}})
$$
where $z=\vert G\vert$, you've written out far more than the product of just the elements of $G$ because a quick sanity check will tell you that $e,g_1,\dots,g_z$, without all the other elements you've included, accounts for $z + 1 > |G|$ many elements, so this is ostensibly not the product you are hoping it is.
Since you've already got the idea, pair an element with its inverse, try this:
$$
\big(\prod_{g\in G}g\big)^2 = \prod_{g\in G}gg^{-1} = e.
$$
(Fill in the missing steps for completeness if you like; they are small.) Hence what can you say about the order of the element $\prod_{g\in G}g$? Finish with Lagrange's theorem.
A: FYI, it is not hard to prove that the product of all the elements of a finite abelian group is always the identity element, except for the case when its Sylow 2-subgroup is (non-trivial) cyclic. Then the product equals the unique element of order $2$.
