Timeline for What is known about the structure of finite groups admitting an automorphism where all elements have "norm" one?

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Feb 18, 2021 at 3:57	comment	added	spin		Some examples if $p$ divides the order of $G$: For $p = 2$, consider $G$ abelian of even order and not elementary abelian. Then $\sigma(g) = g^{-1}$ has order $2$ and $N(g) = e$ for all $g \in G$, but $\sigma$ is not fixed point free. For $p > 2$, consider $G = C_p \times C_p$ and the automorphism $\sigma$ corresponding $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in \operatorname{GL}_2(p)$. Then $A^p = I$ and $I + A + A^2 + \cdots + A^{p-1} = 0$. Thus $\sigma$ has order $p$ and $N(g) = e$ for all $g \in G$, but $\sigma$ is not fixed point free.
Feb 18, 2021 at 0:59	vote	accept	Chris H
Feb 18, 2021 at 10:14
Feb 17, 2021 at 20:27	comment	added	Geoff Robinson		(and fixed by $\sigma$).
Feb 17, 2021 at 18:47	comment	added	Geoff Robinson		That is true, but there will still be an element of prime order $q$ different from $p$.
Feb 17, 2021 at 17:58	comment	added	Thomas Browning		You don't need $q$ to be prime, you just need $q$ to be relatively prime to $p$.
Feb 17, 2021 at 16:36	history	edited	Geoff Robinson	CC BY-SA 4.0	Added new consequence.
Feb 17, 2021 at 16:19	history	edited	Geoff Robinson	CC BY-SA 4.0	typo
Feb 17, 2021 at 16:14	history	edited	Geoff Robinson	CC BY-SA 4.0	typo
Feb 17, 2021 at 13:18	history	answered	Geoff Robinson	CC BY-SA 4.0