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Timeline for answer to On permanents and determinants of finite groups by Gjergji Zaimi

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Sep 1, 2020 at 9:01 history edited darij grinberg CC BY-SA 4.0
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Jul 4, 2020 at 18:11 history edited Gjergji Zaimi CC BY-SA 4.0
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Jul 4, 2020 at 18:06 comment added Gjergji Zaimi @GeoffRobinson Very nice argument!
Jul 3, 2020 at 11:22 comment added Geoff Robinson So the number of $G$ classes in $aH$ is the number of conjugacy classes of $C_{H}(a)$. This is odd because $C_{H}(a)$ has odd order and only the identity is conjugate to its inverse in a group of odd order.
Jul 3, 2020 at 11:19 comment added Geoff Robinson A more group theoretic argument that $aH$ is the union of an odd number of classis is as follows. Notice that $H$ has odd order by assumption, and that we might as well assume that $a$ has order $2$. Then since $\langle a \rangle$ is a Sylow $2$-subgroup of $G$, we see that every element of $G \backslash H$ is $G$-conjugate to an element of the form $ab$ where $b$ has odd order and $ab = ba$. Furthermore, if $b,c \in C_{H}(a)$ have odd order, then $ab$ and $ac$ are $G$-conjugate if and only $b$ and $c$ are already conjugate in $C_{H}(a)$....
Jul 1, 2020 at 15:55 comment added Mare I think I found arxiv.org/abs/1802.08001 showing that the permanent of an elementary abelian 2-group of order $2^n \geq 4$ is indeed nonzero (and divisible by $2^n-1$). Together with your answer, this should completely answer question 2.
Jul 1, 2020 at 14:47 history edited Gjergji Zaimi CC BY-SA 4.0
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Jul 1, 2020 at 5:04 history bounty awarded Mare
Jul 1, 2020 at 3:32 history answered Gjergji Zaimi CC BY-SA 4.0