On almost fixed point free automorphisms (Q1802217)

Suppose that a finite \(p\)-group \(P\) admits an automorphism \(\varphi\) of order \(p^ k\) having exactly \(p^ m\) fixed points. The main result of this paper gives a bound in terms of \(p\), \(k\) and \(m\) (a ``\((p,k,m)\)- bound'') to the derived length of \(P\). The proof uses Krekhin's theorem on Lie rings with fixed-point-free automorphisms of finite order. The author uses and develops also some aspects of the theory of powerful \(p\)- groups (due to Mann and Lubotzky, and to an earlier work of Lazard). The new formula \([M,N]^ p=[M^ p,N]\), for powerfully embedded subgroups \(M\), \(N\), is worth to mention. A bright idea in passing from Lie rings to groups is to use the ``cancellation property'', \(x^{p^ i} \in G^{p^ j} \Rightarrow x \in G^{p^{j-i}}\), enjoyed by uniformly powerful \(p\)-groups. The paper contains also some applications to soluble groups. Earlier, the case of \(k=1\) (i.e. \(| \varphi|=\rho\)) was settled by Alperin using Higman's theorem on Lie rings with fixed-point- free automorphisms of prime order. Reviewer's remark: using some ideas of the present article whose preliminary version was kindly provided by the author, the reviewer proved recently [Mat. Sb. 184, No. 12, 53-64 (1993)] that in the above situation \(P\) has a subgroup of \((p,k,m)\)-bounded index which is soluble of derived length bounded only in terms of \(p^ k\), the order of \(\varphi\). (Earlier, for \(| \varphi|=p\), we also refined Alperin's result cited above, giving a subgroup of \((p,m)\)-bounded index which is nilpotent of \(p\)-bounded class).