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Well be with you, gentlemen. I have the following problem from Aluffi's Algebra: given a finite group $G$ with an unique element $f$ of order $2$, show that \begin{equation} \prod_{g\in G}g=f \end{equation} My reasoning is the following. Since $f$ is the unique element of order $2$, for each $g\in G$ such that $g\neq e$ and $g\neq f$, it must be that $g^{-1}\neq g$. So we can concoct a particular product coupling the elements with their inverses: \begin{equation} e\cdot f\cdot (g_1\cdot g_1^{-1})\cdot(g_2\cdot g_2^{-1})\cdots (g_n\cdot g_n^{-1}) \end{equation} which is really just $f$ in the end. My question is: how can I show that the order of the factors in this product really does not matter? I cannot see a way out.

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    $\begingroup$ If the group is not abelian, what is the meaning of this product? Must it be assumed that the product does not depend on the order? Or that is exactly what is to be proved? $\endgroup$ – ajotatxe Nov 21 '14 at 12:24
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    $\begingroup$ I suspect Aluffi meant to specify an abelian group. It's not true for all ways of ordering a non-abelian group (if you swap the last two elements in $\dots gh$ then you get $\dots hg=\dots gh(h^{-1}g^{-1}hg)$, which is different unless $g$ and $h$ commute, so they can't both be equal to $f$). $\endgroup$ – Jeremy Rickard Nov 21 '14 at 12:28
  • $\begingroup$ So is this question ill-posed by Aluffi and should be revised? $\endgroup$ – Matthew Levy Dec 23 '14 at 19:42
  • $\begingroup$ yes, it seems so. $\endgroup$ – marco trevi Dec 23 '14 at 21:04
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If the group is not abelian, the expression $$\prod_{g\in G} g$$ makes no sense. (It makes no sense at all if the group is infinite, but I assume that it isn't...)

Indeed, take any elements $x,y\in G$ such that $xy\neq yx$ and call the other elements $$g_1,g_2,\ldots,g_r$$ Then, $$g_1g_2\cdots g_rxy\neq g_1g_2\cdots g_ryx$$ so the product of the elements of $G$ depends on the chosen order.

So I think that there are two possibilities: perhaps you must assume that the prod expression makes sense, and you must show that, then, the group must be abelian (as I have just done), or the problem is wrongly set out.

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  • $\begingroup$ The group was finite (I edited the question) but Aluffi did not mention that it was abelian. I should have worked it out myself. Thank you! $\endgroup$ – marco trevi Nov 21 '14 at 13:02
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As noted above, the expression $\prod_{g\in G}g$ does not make sense. To give a concrete example, consider the quaternionic group $G=Q_8=\{\pm1,\pm i,\pm j,\pm k\}$ (under quaternion multiplication). It has only one element of order two, namely $-1$. Now $$ 1\cdot(-1)\cdot i\cdot(-i)\cdot j\cdot(-j)\cdot k\cdot(-k)=-1 $$ but $$ 1\cdot(-1)\cdot i\cdot j\cdot(-i)\cdot(-j)\cdot k\cdot(-k)=1. $$ Therefore one indeed has to specify an order for the product. It is true that there is an order so that the result holds, but it doesn't hold for all orders.

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  • $\begingroup$ Nice example! Good choice of group $\endgroup$ – snulty Nov 21 '14 at 15:38
  • $\begingroup$ @snulty, thanks. $Q_8$ is the smallest nonabelian group with a unique element of order two, and the quaternionic structure makes calculations easy, so the choice was obvious. $\endgroup$ – Joonas Ilmavirta Nov 21 '14 at 16:01
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This is a mistake in textbook. In errata is written that it should be commutative group.

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  • $\begingroup$ Though it is great that you found the errata of the book and have mentioned it here, others had come to the natural conclusion of the group being abelian anyways for the question to make sense. This should have been a comment, but I realize you do not have enough reputation to post it thus. $\endgroup$ – Shailesh Oct 28 '15 at 11:17
  • $\begingroup$ Yes. That is exactly the reason why I put it as an answer. $\endgroup$ – Jiří Nádvorník Oct 28 '15 at 12:40

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