Having a finite $p$-group $G$ ($p$ odd). we denote by $\Omega_1(G)$ the subgroup generated by all the elements of $G$ of order dividing $p$.
Is there an example of such a group $G$, such that $|G:G^p|>|\Omega_1(G)|$?
Assume now that $|G:G^p|>|H:H^p|$, whenever $H<G$. Is it true that $G$ has exponent $p$?
Thanks in advance.
This answer is based on Holt's counter example.
In a $p$-group of maximal class $G$, it is known that:
$|G:G^p|=p^p$ and $\Omega_1(G)$ has either order $p^{p-1}$ or index $p$.
Now we take $G$ of maximal class and order $p^{p+1}$. We claim that if $\Omega_1(G)$ has order $p^{p-1}$ then $G$ is a counterexample to the both questions.
Indeed, the first question is trivial in this case. For the second, every subgroup $H$ of index $>p$ has order $<p^p$ so it satisfies the claim, and if $H$ has index $p$, then $H^p >1$ since otherwise $ H \leq \Omega_1(G)$. Thus $|H:H^p|<|G:G^p|$, whenever $H<G$.
Note also that the two questions are equivalent for such a group. As if $\Omega_1(G)$ has index $p$, then it has order $p^p$, thus it is regular. And as $\Omega_1(G)$ is generated by elements of order $p$, it follows that $\Omega_1(G)$ has exponent $p$. Thus $|G:G^p|=|\Omega_1(G):\Omega_1(G)^p|$.
(One may ask if there is for any $p>2$, a group of maximal class and order $p^{p+1}$ such that $\Omega_1(G)$ is not maximal).
Thanks to Derek Holt.
Answer to the first question: an irregular group $G$ of order $p^{p+1}$ such that $|\Omega_1(G)|=p^{p-1}$ satisfies the condition. In particular, the minimal nonmetacyclic group of order $3^4$, and that group also answers negatively the second equestion. Other examples are:
$G={\rm Q}_8\times{\rm Q}_8$. Here $|G/\mho_1(G)|=2^4>|\Omega_1(G)|=2^2$;
$G={\rm D}_8\times{\rm D}_8$. Here $|G/\mho_1(G)|=2^4<|\Omega_1(G)|=2^6$.
