Let $G$ be a finite group admitting an automorphism $\sigma$ of prime order $p$. Define the norm map $N:G\rightarrow G$ with respect to $\sigma$ by $N(g)= g\sigma(g)\sigma^2(g)\dotsb\sigma^{p-1}(g)$.



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> **Question.** If we know all elements of $G$ have norm equal to $e$, what can be said about the structure of the pair $(G,\sigma)$? In
> particular, must $G$ be nilpotent?
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For instance, if $p=2$, then $G$ must be abelian, but this need not hold for higher primes, e.g., an inner automorphism of a nonabelian $p$ group of exponent $p$.


This condition on an automorphism is a natural weakening of being fixed point free, so its not unreasonable to hope that some of the structural features of that situation hold here.



What is known about the structure of finite groups admitting an automorphism where all elements have "norm" one?
First a note about terminology. It seems in the literature if $\sigma$ is an automorphism of $G$ such that $\sigma^n = 1$ and $g\sigma(g)\sigma^2(g) \cdots \sigma^{n-1}(g) = 1$ for all $g \in G$, we call $\sigma$ a **splitting automorphism**.


After some searching (experts could probably give more detail), I found that the main result of



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> Otto H. Kegel, Die Nilpotenz der $H\_p$-gruppen, Math. Z. 75 (1961), 373–376
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shows that if $G$ is a finite group with a splitting automorphism of prime order, then $G$ must be nilpotent.


The proof by Kegel is based on results of Hughes and Thompson:



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> D. R. Hughes and J. G. Thompson, The $H\_p$-problem and the structure of $H\_p$-groups, Pac. J. Math., 9, 1097–1101 (1959).
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For a finite group $G$ and prime $p$, let $H\_p(G)$ be the subgroup generated by elements of order $\neq p$. Hughes and Thompson show that if $G$ is not a $p$-group, then either $H\_p(G) = 1$, $H\_p(G) = G$, or $[G:H\_p(G)] = p$.


In the case $[G:H\_p(G)] = p$, conjugation by any element $g \in G$, $g \not\in H\_p(G)$ induces a splitting automorphism of order $p$ on $H\_p(G)$.


Conversely, suppose that a group $H$ admits a splitting automorphism $\sigma$ of prime order $p$. Consider the subgroup $G = H \rtimes \langle \sigma \rangle$ of the holomorph of $H$. The definition of a splitting automorphism is precisely that the coset $H\sigma$ consists of elements of order $p$. Then for any integer $k$ coprime to $p$, the coset $H \sigma^k$ also consists of elements of order $p$. This follows from the fact that the map $x \mapsto x^k$ defines a bijection $H\sigma \rightarrow H \sigma^k$ which maps elements of order $p$ to elements of order $p$.


Hence in this case any $g \in G$, $g \not\in H$ has order $p$, so $H\_p(G) \leq H$. If $H$ is not a $p$-group, then $H\_p(G) \neq 1$ and by the result of Hughes and Thompson we must have $H\_p(G) = H$.


So for finite groups which are not $p$-groups, the study of splitting automorphisms of order $p$ comes down to the study of what Hughes and Thompson call "$H\_p$-groups", i.e. groups of the form $H\_p(G)$, where $G$ is a finite group such that $[G:H\_p(G)] = p$. This is seen as a generalization of Frobenius groups, which relate to fixed point free automorphisms of order $p$.


Hughes and Thompson show that a $p$-Sylow of an $H\_p$-group has a normal complement which is nilpotent, so in particular $H\_p$-groups are solvable. In his paper Kegel continues from there, and shows that $H\_p$-groups are nilpotent.




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I note the following about the relation between "fixed point free" and "splitting automorphism".


An argument from (Gorenstein, *Finite groups*, Lemma 1.1 in Chapter 10): suppose that $\sigma$ is a fixed point free automorphism of $G$ of order $n$. Then the mapping $x \mapsto x^{-1}\sigma(x)$ is injective, hence also surjective. Writing $g \in G$ as $g = x^{-1}\sigma(x)$, we see that $g \sigma(g) \cdots \sigma^{n-1}(g) = e$.


So $\sigma$ being fixed point free implies that $\sigma$ is a splitting automorphism.


The converse fails. For $p = 2$, consider $G$ abelian of even order and not elementary abelian. Then $\sigma(g) = g^{-1}$ clearly defines a splitting automorphism of order $2$, but $\sigma$ is not fixed point free. For $p > 2$, consider $G = C\_p \times C\_p$ and the automorphism $\sigma$ corresponding to $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in \operatorname{GL}\_2(p)$. Then $A^p = I$ and $I + A + A^2 + \cdots + A^{p-1} = 0$. Thus $\sigma$ is a splitting automorphism of order $p$, but $\sigma$ is not fixed point free.


