Math-1188 · mathematics · Group theory

Math-1188

Kourovka Notebook Problem 21.1

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Statement

Definitions and mathematical statement

For a positive integer $n$, a finite group $K$ and an automorphism $\varphi$ with $\varphi^n=\mathrm{id}$, use $x^{\varphi^j}=\varphi^j(x)$ and define the twisted solution set \[ X_{n,\varphi}(K)=\{x\in K:x\varphi(x)\cdots\varphi^{n-1}(x)=1\}. \] The factors are multiplied in the displayed order. Define $c_n$ as the supremum of $|X_{n,\varphi}(K)|/|K|$ over all these pairs for which $X_{n,\varphi}(K)\ne K$. Thus $c_n<1$ asks for one positive gap from one, uniform over all finite groups and admissible automorphisms, after excluding pairs where the equation holds identically.

Part (a) asks: for every $n>1$, does the hypothesis $c_{p^a}<1$ for every prime power $p^a$ dividing $n$, with $a\ge1$, imply $c_n<1$? The bounds for different prime powers need not be identical.

For part (b), put \[ H_n(G)=\langle\{x\in G:x^n\ne1\}\rangle \] for each finite group $G$, where brackets denote the subgroup generated by the indicated elements; the empty set generates $1$. Suppose there exists an integer $k_n\ge1$ depending only on $n$ such that $[G:H_n(G)]\le k_n$ whenever $H_n(G)\ne1$. Does this uniform index hypothesis imply $c_n<1$?

The index condition in (b) involves ordinary powers and arbitrary finite $G$, whereas its conclusion controls twisted powers for arbitrary $\varphi$. Both parts ask implications for a fixed integer parameter, not bounds depending on the individual group.

Short English statement

Do prime-power density gaps, or a uniform generalized Hughes–Thompson index bound, force a uniform gap for twisted $n$th-power solutions?

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