Timeline for The parity of the full automorphism group order of finite non-abelian groups of prime exponent

7 events

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Jan 26, 2018 at 0:22	history	edited	YCor	CC BY-SA 3.0	Added comments distinguishing $p=3$ and $p=5$.
Jan 20, 2016 at 10:22	comment	added	YCor		Yes you really have to use BCH. You have a presentation with generators $x_1,\dots,x_7$, relators $x_i^7=1$ and, for $i<j$, $x_ix_jx_i^{-1}x_j^{-1}=m_{ij}$ for all $i,j$, where $m_{ij}$ is a word in $x_{j+1},\dots,x_7$. With BCH you can compute $m_{ij}$ in the Lie algebra, that is some combination $(\ell(j,j+1),\dots,\ell(j,7))$. This has to be converted into a word of the form $x_{j+1}^{k(j,j+1)}\dots x_7^{k(j,7)}$. Then $k(j,j+1)=\ell(j,j+1)$. To find the next exponents, first compute $x_{j+1}^{-k(j,j+1)}m_{ij}$; then $k(j,j+2)$ is the $j+2$-coefficient in the resulting element, and so on.
Jan 20, 2016 at 6:27	comment	added	Alireza Abdollahi		Is it possible to give an explicit presentation for the group of order $p^7$ in your last EDIT? I have used the same relations as the Lie algebra with the relations of 7th powers of all generators. This gives me a group of nilpotency class 5, expoenent 7 and order $7^6$ whose automorphism group is of order 242121642. Is this because my group is not obtained from BCH?
Jan 19, 2016 at 12:44	history	edited	YCor	CC BY-SA 3.0	added 901 characters in body
Jan 19, 2016 at 12:23	vote	accept	Alireza Abdollahi
Jan 18, 2016 at 13:06	history	edited	YCor	CC BY-SA 3.0	fixed issue for small $p$ in the last paragraph
Jan 18, 2016 at 12:06	history	answered	YCor	CC BY-SA 3.0