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Back when I was an undergraduate we were asked to name a finite noncommutative group such that more then half the elements have order two. Furthermore, we know that every group such that every element different from the identity has order two must be abelian. So I ask myself, "how much of an noncommutative group can be of order two". In a more rigorous way:

What do I get, when I take the supremum over all finite noncommutative groups $G$ of the following expression $$ \frac{\vert\{ g\in G \ : \ g \text{ has order } 2\}\vert}{\vert G \vert}.$$

I was able to show that the supremum should be greater equal $2/3$ by considering the groups $G_n =(\mathbb{Z}/2\mathbb{Z})^n \oplus S_3$ which has $4\cdot 2^n -1$ elements of order two and $\vert G_n \vert = 6\cdot 2^n$.

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    $\begingroup$ A similar trick with $\mathbb Z_2^n\oplus D_4$ will give you $3\over4$. I believe the final answer to be 1, but that's merely a guess. $\endgroup$ – Ivan Neretin May 10 '18 at 22:00
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    $\begingroup$ Looks like the definitive part of my comment was right, but my guess was wrong. $\endgroup$ – Ivan Neretin May 10 '18 at 22:23
  • $\begingroup$ @IvanNeretin I would also have guessed that the ration goes to 1. $\endgroup$ – Severin Schraven May 11 '18 at 13:22
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This preprint has a self-contained proof that the answer is $\frac34$:

Allan L. Edmonds and Zachary B. Norwood (2009) "Finite groups with many involutions" https://arxiv.org/abs/0911.1154

It is shown that a finite group in which more than 3/4 of the elements are involutions must be an elementary abelian 2-group. A group in which exactly 3/4 of the elements are involutions is characterized as the direct product of the dihedral group of order 8 with an elementary abelian 2-group.

(I'll leave this answer as community wiki in case someone else wants to read the paper and summarize the argument.)

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  • $\begingroup$ Cool, thank you very much. $\endgroup$ – Severin Schraven May 11 '18 at 13:21

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