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?



Bound for the Frattini subgroup of a $p$-group
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 $[G: \Phi(G)] = p^{2}.$


