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For $H$ a group and $n\in\mathbb{N}$, let $H^{(n)}=\langle h^n : h \in H \rangle$. Now let $G$ be an extraspecial $p$-group (see definition). Is it true that $G^{(p)}\cong \mathbb{Z_p}$. (It holds for $D_8$ and $Q_8$.)

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No. It is false for the extraspecial p-groups of exponent p, for all odd p. In particular, it is false for the Sylow p-subgroup of GL(3,p).

In general G(p) ≤ Φ(G), but you need not have equality when p is odd.

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Thanks a lot! I have some further questions if you'd like (see below). – Bart Patzer May 26 '11 at 7:54

For odd $p$, the extraspecial $p$-groups are central products of the nonabelian group of order $p^3$ and exponent $p$ and at most one copy of the nonabelian group of order $p^3$ and exponent $p^2$; that is, if $$\begin{align*} G &= \langle x,y,z\mid x^p=y^p=z^p=[x,z]=[y,z]=1, [x,y]=z\rangle,\\ H &= \langle x,y\mid x^{p^2}=y^p=1, yx = x^{p+1}y\rangle \end{align*}$$ then every extraspecial $p$-group is obtained by taking a direct product of finitely many copies of $G$, at most one copy of $H$, and then identifying the commutator subgroups together.

Since the group is of small class, it is regular, so $E^{(p)}=E^p$, the set of $p$th powers of $E$. If $E$ consists only of copies of $G$, then $E^p$ is trivial; otherwise (if it contains a copy of $H$), then $E^p = [E,E]=Z(E)\cong \mathbb{Z}_p$.

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Thanks! Since there is exactly two extraspecial groups of order $p^{2n+1}$, am I right to assume that when $p$ is odd then one of these satisfy $G^{(p)}$ and the other does not? What do you mean by "identifying the commutator subgroups togerther"? Do you know either a book or article where the above presentations of $G$ and $H$ are to be found (for reference). – Bart Patzer May 26 '11 at 7:53
(1) Yes: $G^{(p)} = 1 < \Phi(G)$, $H^{(p)} = \Phi(H) \cong \mathbb{Z}_p$. Similarly for the larger groups. (2) Take generators $x_1, \dots, x_n, y_1, \dots, y_n, z$ where for each $i$, $x_i, y_i, z$ define one of the groups $G$ or $H$, and where $x_i,y_i$ both commute with $x_j,y_j$ when $i \neq j$. "Identify the commutator subgroups" means instead of $z_i$ we just have one $z$. (3) Daniel Gorenstein's Finite Groups is a good place to learn about important $p$-groups. – Jack Schmidt May 26 '11 at 11:44
@Moishe: As Jack says. Or alternatively, take $\times G_i$, where each $G_i\cong G$, and then mod out by the subgroup generated by all central elements of the form $z_iz_j^{-1}$, which as the effect of making the commutator subgroups of each copy "the same". – Arturo Magidin May 26 '11 at 15:41
Thanks again, I really appreciate the help. – Bart Patzer May 26 '11 at 19:22

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