Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Assume that $G$ is a finite $p$-group, $p$ odd, with a non-trivial elementary abelian Frattini subgroup. Then both $\Phi(G)$ and $G/ \Phi(G)$ are vector spaces over $\mathbb{F}_p$. Is it possible to get a bound for $\dim \Phi(G)$ as a (polynomial) function of $\dim G / \Phi(G)$?

Edit: I should probably write a little more. First of all, the exponent of such a group is $p$ or $p^2$ as the $p^{\text{th}}$ power of any element falls in the Frattini subgroup. If the exponent of the group is $p$ then $\Phi(G) = G'$, since $\Phi(G)=G^pG'$ in general. Now suppose that $g_1,\dots,g_n$ is a basis for $G / \Phi(G)$. Then it is easy to see that the elements $[g_i,g_j]$, $0 \leq i,j \leq n$ generate $G'$ because, in this particular case, the $p^{\text{th}}$ power map and the commutator map (with fixed first or second coordinate) are homomorphisms. So if the exponent of $G$ is $p$ then $\dim \Phi(G)$ is at most $\dim G / \Phi(G) \choose 2$. My guess is that there are $p$-groups which achieve this upper bound.

What happens when the exponent of the group is $p^2$? I guess what I'm really asking is whether the upper bound remains quadratic in that case too.

The answer to that question is $\textbf{no}$, as Geoff's post establishes.

Is there an upper bound that takes into account $p$ as well?

share|improve this question
3  
Yes, there are $p$-groups that achieve the bound for exponent $p$; namely, the relatively free groups of rank $n$, class $2$, and exponent $p$ have commutator subgroup that is free abelian of rank $\binom{n}{2}$; this group can be realized as $F_n/F_n^p(F_n)_3$, where $F_n$ is the absolutely free group of rank $n$, and $(F_n)_3$ is the third term of the lower central series of $F_n$. –  Arturo Magidin Apr 19 at 0:18

2 Answers 2

up vote 6 down vote accepted

There can be no polynomial bound which is independent of $p$. Consider the group $G = C_{p} \wr C_{p},$ where $C_{p}$ denotes the cyclic group of order $p.$ Then $\Phi(G) = G^{\prime}$ has order $p^{p-1},$ (and is elementary Abelian) yet $[G: \Phi(G)] = p^{2}.$

share|improve this answer

For your second question the answer is yes. The bound follows from Schreier's inequality: if $\Phi(G)$ has index $p^d$, then it follows that $\Phi(G)$ can be generated by $p^d(d-1)+1$ elements. Note that this is true without assuming that the Frattini subgroup is elementary abelian.

Your first claim is true for the exponent $p^2$ case, if one requires that $\Phi(G)$ is central. Once you fix the number of generators $d$, the largest such group can be defined to be the quotient of the free group $F$ on $d$ generators by the subgroup $[F,F^p[F,F]] (F^p[F,F])^p$. In that case the Frattini subgroup can be generated by $d(d+1)/2$ elements.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.