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In Aluffi's Algebra Chapter 0, there's an error in the 2009 publication that took me a while to find:

Let $G$ be a finite abelian group with exactly one element $f$ of order 2. Prove that $\prod_{g\in G}g = f$.

With the addition of the abelian condition, the problem is much simpler. However, I could not find a counterexample or proof for the non-abelian case. How can I go about proving/finding a counterexample to the statement without the abelian condition?

Shaun
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perpetuallyconfused
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    Without the abelian condition, the problem is not quite well-formulated. What order should the product be in? – Tobias Kildetoft Apr 13 '20 at 13:49
  • Try the quaternion group of order $8$. – lulu Apr 13 '20 at 13:50
  • @TobiasKildetoft ah yes, I had struggled with that at first but then forgot about it, that would resolve the question then wouldn't it – perpetuallyconfused Apr 13 '20 at 13:51
  • @lulu how did you come up with $Q_8$? One of the big issues that I was having was an inability to think of nonabelian groups that fit the hypotheses – perpetuallyconfused Apr 13 '20 at 13:56
  • Some notes on groups with this property: The Sylow $2$-subgroup must be normal, and either be cyclic or generalized quaternion. So indeed, the simplest example is $Q_8$. – Tobias Kildetoft Apr 13 '20 at 14:01
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    You know from Sylow theory that the unique element of order $2$ must be central. $Q_8$ is the smallest candidate. – lulu Apr 13 '20 at 14:03
  • See also https://math.stackexchange.com/questions/1353829/product-of-all-elements-in-finite-group – lhf Apr 13 '20 at 16:07

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Consider the quaternion group of order $8$, $Q_8$.

That has a unique element of order $2$, namely $-1$.

Of course there is an ordering for the product that gives the desired result but we could also have the ordering:

$$-1\times i\times j\times -i\times k\times -k\times -j=1$$

lulu
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