Question about order of product of elements in a group

Question

Let G be a finite abelian group. Prove that the product of all elements in G has order 2. I think i am supposed to use lagrange's theorem but how?

Answer 1

Call an element $a$ of the group lonely if $a$ is its own inverse. The group then consists of a set $L$ of lonely elements $l_1,\dots,l_k$, and a set $P$ of pairs of distinct elements $x,y$ such that $xy=1$.

The product of all the elements of the group is equal to $l_1l_2\cdots l_k$.

Note that $(l_1l_2\cdots l_k)^2=1$.

Thus the product of the elements of the group has order $2$ or $1$. The order is not necessarily $2$. (Look at the $1$-element group.)

Answer 2

Well I think you mean "has order $\leq 2$"

Take $C_5 = \{ 1, x, x^2, x^3, x^4 \}$

Clearly the product is $x^{10} = 1$, that has order $1$

This estabilished, the product of all elements contains $x$ and $x^{-1}$ $\iff$ $x\neq x^{-1}$. So the product of all elements is the product of all elements of order $2$, since the others cancel. And that element has still order $2$ because $G$ is abelian.