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I have known cases of abelian and nonabelian groups that have a center of order p but never a power of p. Is there at least a known case where $|Z(G)|=p^2$?

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How about an abelian group of order $p^2$? If you want non-abelian, you could take the direct product of two non-abelian groups each of which have center of order $p$. – Mikko Korhonen Sep 14 '12 at 8:00
So what's the easiest nonabelian example which is not a direct product? How about the group of upper untriangular $3 \times 3$ matrices over the finite field of order $p^2$? – Derek Holt Sep 14 '12 at 8:13
You could take a symmetric group $S_n$, which has trivial centre for $n>2$, and is nonabelian, and take a direct product with a group of order $p^2$ (which is necessarily abelian). – Mark Bennet Sep 14 '12 at 8:31
Keep a lookout for groups with center the Klein $V$ group. They come up a lot. – Alexander Gruber Oct 16 '12 at 21:49

In what follows, we construct a $p$-group with centre of order $p^n$ for arbitrary $n\geq 1$ (in fact, we constructs groups with arbitrary centre, but the construction gives a $p$-group if the centre has order $p^n$). It is (essentially) Exercise 5.2.2. from Robinson's book "A Course in the Theory of Groups" (p138 in my version).

For your example, you want to take $A=C_{p^2}$ or $A=C_p\times C_p$. These two choices of $A$ will give two non-isomorphic non-abelian $p$-groups with centre of order $p^2$.

Let $A$ be a non-trivial abelian group and set $D=A\times A$. Define $\delta\in\operatorname{Aut}(D)$ as $(a_1, a_2)^{\delta}=(a_1, a_1a_2)$. Let $G$ be the semidirect product $\langle \delta\rangle\ltimes D$.

Then one can prove that this group is finite if $A$ is finite, and moreover (part $(a)$ of the exercises) that $Z(G)=G^{\,\prime}\cong A$.

Parts $(b)$ and $(c)$ of the question are,

$(b)$ Prove that $G$ is a torsion group if and only if $A$ has finite exponent (note that finite exponent is stronger than just torsion),

$(c)$ Deduce that even if the centre of a nilpotent group is a torsion group, the group may contain elements of infinite order,

which are entirely irrelevant to the question at hand, but interesting nonetheless.

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(c) is especially weird since the same thing in reverse is true: if $G/[G,G]$ is torsion, then $G$ is torsion (or not nilpotent). The upper central version requires finite exponent, but the lower central does not (as far as I can see). – Jack Schmidt Sep 14 '12 at 13:22

$\mathbb{Z} / p^2 \mathbb{Z}$, the cyclic group of order $p^2$, is an abelian group of order $p^2$. Hence $|Z(G)| = p^2$.

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That's nice. Keep the examples flowing guys. I want to see a nonabelian one too – Student Sep 14 '12 at 8:03

Certainly any abelian group of order $p^2$ does. The two such abelian groups are (up to isomorphism) $\mathbb Z_{p^2}$ and $\mathbb Z_p\times \mathbb Z_p$.

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Take any simple group $\,G\,$ and direct-multiply it with an abelian group of order $\,p^2\,$ , say $\,H:=G\times C_{p^2}\,$ , and here you have a non-abelian example.

Of course, instead of simple above you can take any centerless groups, say $\,S_n\,\,,\,\,n\geq 3\,$ . This is what Mark commented before.

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