MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let denote by $V = V(c, s, t)$ the class of all finite $p$-groups such that $class(G) \leq c$, $expZ(G) \leq p^s$, and $G/Z(G)$ has a maximal normal abelian subgroup of rank $\leq t$.

The class $V$ contains infinitely many $p$-groups since it contains $G \times (Z/(p))^k$, whenever $G$ is contained in $V$.

Is it true that $V$ contains infinitely many p-groups without an abelian diraect factor?

More particularly, is it true that $V$ contains infinitely many p-groups $G$ in which the center is contained in the frattini subgroup?

share|cite|improve this question
Ok, I think that I have an answer. A p-group G in the class V has exponent at most p^s. Also if $Z(G) \leq \Phi (G)$ then d(G) the minimal number of generators of G equals to $d=d(G/Z(G)$, and by a nice result of Thompson (A. Mann for 2-groups), d can not exceed $t(t+1)/2$. It follows from Zelmanov solution of the restricted Burnside problem that V contains only finitely many p-groups with a center contained in the frattini subgroup. Also It seems to me that a variation of the above argument answers the first question negatively. – Yassine Guerboussa Jun 19 '13 at 16:58
in the first line of the above comment, I mean "...exponent at most $p^{sc}$". – Yassine Guerboussa Jun 19 '13 at 17:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.