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 $Q_8$.)
$p$-th powers of elements of an extraspecial $p$-group
2 Answers
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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0Thanks a lot! I have some further questions if you'd like (see below). – 2011-05-26
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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0@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". – 2011-05-26