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Is it true that if $G$ is a $p$-group, where $p$ is a prime, then center of $G$ is non-trivial?

I know for finite $p$-group center of $G$ is non-trivial (easy to prove using class equation for group). But I am not sure about infinite $p$-group. I have no idea how to approach this problem. I know Prufer group is an example of infinite $p$-group. Is the center of Prufer group is trivial?

Any help would be appreciated...

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    $\begingroup$ The Prufer group, being abelian, is its own center. I would think that the center of the free product $\Bbb Z_p* \Bbb Z_p$ is trivial (edit: the free product is not a $p$ group). $\endgroup$ – Quang Hoang Oct 27 '14 at 15:28
  • $\begingroup$ @QuangHoang could you please explain me why center of the free product $Z_p*Z_p$ is trivial. $\endgroup$ – Ripan Saha Oct 27 '14 at 15:32
  • $\begingroup$ Intuitively, free product doesn't have any relations except for the internal relations of the component groups. Therefore you can't have any element that commutes with both elements in the two components. But then again, the free product is not a $p$-group. $\endgroup$ – Quang Hoang Oct 27 '14 at 15:36
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    $\begingroup$ There's a particularly nice construction here, and some further discussion here. $\endgroup$ – Mike Earnest Oct 27 '14 at 16:02
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The center of an infinite $p$-group can be trivial. From Rotman's An Introduction to the Theory of Groups, p. 115:

... there is an example of McLain (1954) of an infinite $p$-group $G$ with $Z(G)=1$, with $G'=G$ (so that $G$ is not even solvable), and with no characteristic subgroups other than $G$ and $1$.

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Tarski monsters are finitely generated simple $p$-groups for $p>10^{75}$. Every proper subgroup is cyclic of order $p$, in particular. There is a continuum of non-isomorphic such groups for each such $p$.

The moral of the story is: infinite groups have a lot of crazy possibilities, and that's why you you see all these special classes of specific kinds of infinite groups.

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