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?

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    $\begingroup$ 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 $F_n$. $\endgroup$ Commented Apr 19, 2014 at 0:18

2 Answers 2


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 $[G: \Phi(G)] = p^{2}.$


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.


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