Let $G$ be a finite group of order $o(G)$ and containing $k(G)$ conjugacy classes. It has been shown by P. M. Neumann [Bull. London Math. Soc. 21 (1989), no. 5, 456--458; MR1005821 (90f:20036)] that if $k(G)ge o(G)/A, A$ some positive number, then $G$ contains two normal subgroups $Kle H$ with $H/K$ abelian, $[G:H]le A$ and $K)le f(A)$, for a suitable function $f$ of $A$ (whose order of magnitude is $A^A)$. In this note the authors restrict themselves to the consideration of $p$-groups and characterization of those for which $A$ is relatively small. For example, it is shown (Theorem 1) that if the $p$-group $G$ has $k(G)o(G)/p^3$ then at least one of the following holds: (i) $o(G_2)le p^2$, where $G_2$ is term 2 of the lower central series of $G$; (ii) $o(G)/Z(G))le p^3, Z(G)$ the center of $G$; or (iii) $G$ has an abelian maximal subgroup $M$. It is noted that (i) or (ii) both imply that no element of $G$ has more than $p^2$ conjugates while (iii) implies that $M$ contains at least $p^n-2 G$-conjugacy classes, $o(G)=p^n$. Thus Theorem 1 has a converse when $p$ is odd and yields the following (Corollary): If $G$ is an odd order $p$-group then it has breadth 2 if and only if $o(G_2)=p^2$ or $o(G/Z(G))=p^3$.
Groups of Prime Power Order with Many Conjugacy Classes
GAVIOLI, NORBERTO;SCOPPOLA, CARLO MARIA
1998-01-01
Abstract
Let $G$ be a finite group of order $o(G)$ and containing $k(G)$ conjugacy classes. It has been shown by P. M. Neumann [Bull. London Math. Soc. 21 (1989), no. 5, 456--458; MR1005821 (90f:20036)] that if $k(G)ge o(G)/A, A$ some positive number, then $G$ contains two normal subgroups $Kle H$ with $H/K$ abelian, $[G:H]le A$ and $K)le f(A)$, for a suitable function $f$ of $A$ (whose order of magnitude is $A^A)$. In this note the authors restrict themselves to the consideration of $p$-groups and characterization of those for which $A$ is relatively small. For example, it is shown (Theorem 1) that if the $p$-group $G$ has $k(G)o(G)/p^3$ then at least one of the following holds: (i) $o(G_2)le p^2$, where $G_2$ is term 2 of the lower central series of $G$; (ii) $o(G)/Z(G))le p^3, Z(G)$ the center of $G$; or (iii) $G$ has an abelian maximal subgroup $M$. It is noted that (i) or (ii) both imply that no element of $G$ has more than $p^2$ conjugates while (iii) implies that $M$ contains at least $p^n-2 G$-conjugacy classes, $o(G)=p^n$. Thus Theorem 1 has a converse when $p$ is odd and yields the following (Corollary): If $G$ is an odd order $p$-group then it has breadth 2 if and only if $o(G_2)=p^2$ or $o(G/Z(G))=p^3$.Pubblicazioni consigliate
I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.