Timeline for answer to On permanents and determinants of finite groups by Gjergji Zaimi
Current License: CC BY-SA 4.0
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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 |