# Finitely generated infinite $p$-group with a unique subgroup of order $p$

I'd like to ask can we characterize the structure of finitely generated infinite $p$-group which has a unique subgroup of order $p$?

Can we say that these group are residually nilpotent?

Any comments are welcome.

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What's your favorite example of such a group? – PseudoNeo Jul 11 '12 at 22:33
The unique subgroup of order p is normal (being unique) and is contained in every cyclic subgroup (since it is the unique cyclic subgroup of order p), and so every non-identity normal subgroup contains this subgroup. In particular, such a group is residually nilpotent if and only if it is nilpotent. – Jack Schmidt Aug 28 '12 at 19:43
Let $T_i=\langle a, b; r_1, r_2, \ldots \rangle$ be a Tarski monster (every subgroup has order $p$ for a fixed prime $p$). Ol'shankii proved that there are a continuum of such groups. Then, let $S_i=\langle a, b, x; x^p, [a, x], [b, x], r_1=x, r_2=x, \ldots\rangle$. If $W(a, b)=_{T_i}1$ then $W(a, b)=_{S_i}x$ and so $U(a, b)$ has order $p^2$ for all words $U$ over $a$ and $b$. Moreover, $V(a, b, x)$ has the form $U(a, b)x^i$ and so this has order $p^2$ (unless $V$ is the empty word). My point is, as there is a continuum of Tarskii monsters there is a continuum of the groups you are looking at. – user1729 Aug 29 '12 at 9:57
(You still need to prove pairwise isomorphic, but just quotient out the unique subgroup of order $p$ and you're done!) – user1729 Aug 29 '12 at 10:00
Every finite subgroup of such a group should be cyclic. Also this group has a non-trivial center. Have you an example of such a group that is not abelian? – Yassine Guerboussa Nov 19 '13 at 12:09