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Let $G$ be a finite group and $n > 1$ a divisor of $|G|$. Let $P_n(G)$ be the subgroup of $G$ which is generated by all elements which satisfy $x^n = 1$. Since $n > 1$, by Cauchy's theorem we have $P_n(G) > 1$. It is also easy to see that $P_n(G)$ is a characteristic subgroup of $G$. Hence if $G$ is characteristically simple (ie. direct product of isomorphic simple groups), then $P_n(G) = G$ for all $n > 1$, $n$ divisor of $|G|$ . There are plenty of other examples in finite $p$-groups, dihedral group $D_8$ also has this property.

My question: Is there a finite group $G$ that is not characteristically simple or a $p$-group, but for which $P_n(G) = G$ for every divisor $n > 1$ of $|G|$?

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If $d|n$ and $d>1$ then $P_d(G)\subseteq P_n(G)$. So to prove true for all $n$, we only need to show for $n$ prime. –  Thomas Andrews Nov 28 '12 at 19:33
    
If $G=P_n(G)$, then every Frattini cover $H$ of $G$ also satisfies $H=P_n(H)$. –  Lior B-S Nov 29 '12 at 12:37

1 Answer 1

up vote 4 down vote accepted

Yep. Here are a few examples which hold this property:

  • The Valentiner group (triple cover of $A_6$), which has a center of order $3$.

  • The inner automorphism group of $\mathbb{Z}_2\wr A_5$, which has a characteristic subgroup of order $16$.

  • The Frattini extension of $PSL(3,2)$ by $(\mathbb{Z}_2)^3$.

EDIT: I did some computational experiments looking for a simpler example and found that the second group listed above is the smallest order example which holds this property (tied with some other unnamed group). Furthermore, there only seem to be around $6$ such groups of order less than $2000$. So, I guess these are pretty uncommon.

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