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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.

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    $\begingroup$ Try SmallGroup(81,10) for both questions. $\endgroup$
    – Derek Holt
    Feb 1, 2014 at 16:45
  • $\begingroup$ "agemo"? "omega" spelled backwards? $\endgroup$ Feb 2, 2014 at 12:12
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    $\begingroup$ @GerryMyerson: Yes, the name comes from the common notation $\mho_1(G) = G^p$. $\endgroup$ Feb 2, 2014 at 12:21

2 Answers 2

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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.

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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:

  1. $G={\rm Q}_8\times{\rm Q}_8$. Here $|G/\mho_1(G)|=2^4>|\Omega_1(G)|=2^2$;

  2. $G={\rm D}_8\times{\rm D}_8$. Here $|G/\mho_1(G)|=2^4<|\Omega_1(G)|=2^6$.

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