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!

  • $\begingroup$ Clearly, the group $G$ is from even order. $\endgroup$ – mrs Nov 15 '12 at 11:57
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    $\begingroup$ It's not true for arbitrary orders of the group elements. For example, in the quaternion group $Q_8$ the product can be either $f$ or the identity. $\endgroup$ – Derek Holt Nov 15 '12 at 12:06
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    $\begingroup$ You need to change it to finite abelian group. See math.fsu.edu/~aluffi/algebraerrata/Errata.html (p.49) $\endgroup$ – wj32 Nov 15 '12 at 12:10
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    $\begingroup$ (The author even says: "My personal favorite is the missing abelian at p.49, Exercise 1.8.") $\endgroup$ – wj32 Nov 15 '12 at 12:16
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    $\begingroup$ @BabakSorouh, let $\,G=\{g_0:=1,g_1,...,g_n\}\,\,,\,n\,$ an even natural, be a group. In the quotient group $\,G/G'\,$ we have that the factors of the product $$\prod_{i=0}^n(g_iG')$$ are commutative, so we can pair each coset $\,g_iG'\,$ with $\,g_i^{-1}G\,$ and we thus get... $\endgroup$ – DonAntonio Nov 16 '12 at 3:06

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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    $\begingroup$ You seem to be assuming $\,G\,$ is abelian, which is not given. $\endgroup$ – DonAntonio Nov 15 '12 at 12:33
  • $\begingroup$ You have a word of elements of $G$ and cannot use that property without assuming $G$ is abelian, so I agree with Don. $\endgroup$ – mrs Nov 15 '12 at 12:36
  • $\begingroup$ Of course, you are right. I apologize. $\endgroup$ – Espen Nielsen Nov 15 '12 at 12:55
  • $\begingroup$ The question is, do we have $|abcb^{-1}d| = |acd|$ in a non-abelian group? Because if we do, then your result holds; you can remove pairs $(b, b^{-1})$ from the word without changing the total order. I don't know, but by conjugation by $d$ above, we can rename $da$ to $a$ and explore whether $|abcb^{-1}| = |ac|$ $\endgroup$ – Arthur Nov 15 '12 at 13:07
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    $\begingroup$ It's clear that in any nonabelian group you can get at least two different products of all group elements, by using different orders. $\endgroup$ – Derek Holt Nov 15 '12 at 13:33

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