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I am doing some exercises in Algebra: Chapter 0. In the second chapter, we are asked to prove the following:

$G$ is a finite group with a unique element $f$ of order $2$. Then $\operatorname{\Pi_{g\in G}}g=f$.

This result is highly plausible. If we multiply the elements in the order of \begin{equation}e\cdot f\cdot \text{elements of order 3}\cdot\text{elements of order 4}\cdots,\end{equation} and pair elements with their inverses, then we get $f$, since it is the only element that does not have a couple.

However this is only one possible order of multiplication, and we know that in general different order give different results.

So I wonder how we can do the general case. Thanks!

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    @BabakSorouh, knock yourself out. Thanks.2012-11-16

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In general we have a result about the order of elements that says $|aba^{-1}|=|b|$ for all $a,b\in G$.

Let $p=\left|\prod_{g\in G} g\right|$

Using this result, assuming $G$ is abelian, we can steadily "remove" pairs $(a,a^{-1})$ from $p$. However, $f$ is the only element which is its own inverse, so this process stops when we have $p=|f|=2$. But $f$ is the only element of order $2$ in $G$, so

$\prod_{g\in G} g = f$

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    Thanks! But for abelian groups it is trivial, and actually my problem already contains the idea of the proof.2012-11-15