A persisting element in all subgroups.

Question

Let $G$ be a finitely generated abelian group and $a$ be a nontrival element of $G$ contained in all nontrivial subgroups of $G$. Is $G$ necessarily cyclic?

Answer 1

Let's consider what is the condition such that $G$ would have such element $a$. For simplicity, only finite $G$ are examined.

First, $|G|$ must be a prime power, otherwise it is false by Cauchy's. Let $|G|=p^{n}$ ($p$ prime, $n\geq 1$). Then any abelian subgroup of $G$ must be cyclic, by the proof using cyclic decomposition as outline in the comment: for any nontrivial abelian subgroup $H$ of $G$, we can always do the cyclic decomposition $H=\mathbb{Z}_{k_{1}}\times \mathbb{Z}_{k_{2}}\times\ldots\mathbb{Z}_{k_{t}}$. If $t\geq 2$ then $H$ have 2 subgroups that intersect trivially: $\mathbb{Z}_{k_{1}}$ and $\mathbb{Z}_{k_{2}}$ (or more correctly, their embedding in $H$), which in turn mean that they are both subgroup of $G$ that not both containing $a$. Hence $t=1$ so $H$ is cyclic.

By a theorem, http://books.google.com/books?id=0268b52ghcsC&pg=262#v=onepage&q&f=false all subgroup of a such group must be either cyclic or a generalized quarternion group. In particular, $G$ is either cyclic or a generalized quarternion group. Hence the only counterexample would be the generalized quarternion group:

$<x,y|x^{2n}=y^{4}=1,x^{n}=y^{2},y^{-1}xy=x^{-1}>$

Because $|G|=p^{n}$ this means $p=2$. Generalized quarternion group have order at least $8$. Hence $G$ is a generalized quarternion group with order $2^{n}$ for $n\geq 3$ are all the exception to the claim. Note that the only possible persistent element in those counterexample are the lone element with order $2$.

As for infinite $G$, I have no ideas. Note that every cyclic group must have finite order, as infinite cyclic group do not have persistant element. Also, just like above, we still have $a$ have order $p$ for some prime $p$, and so each element still have order a power of $p$. Hence $G$ is now an infinite $p$-group that is finitely generated. The existence of such a group is the famous nontrivial Burnside's problem. There is a theorem that prove such a group exist http://en.wikipedia.org/wiki/Golod-Shafarevich_theorem but there is no guarantee (and in fact rather unlikely) that the group constructed by that theorem will have a persistent element. Hence I will leave this to the expert.

Now if we remove the requirement that the group have to be finitely generated, then an example can be found. Notice that you can embedded $Q_{4n}$ inside $Q_{4(2n)}$ (the generalized quarternion group). Hence by keep embedding like that, you get a chain of group (chain ordered by subset relation). Simply use take the union of the whole chain to get the infinite group you want.

Answer 2

The group $G$ is supposed to be abelian. The subgroup generated by the persistent element must be simple. The group can't have any other simple subgroups, because the sum of simple subgroups is a direct sum of simple groups. Also the group must be indecomposable, because the persistent element can't belong to both a direct summand and its complement.

If $H$ is the simple subgroup generated by the persistent element, it intersects non trivially every non zero subgroup, so it is essential. In particular, the injective envelope of $G$ is the same as the injective envelope of $H$, so it is the Prüfer $p$-group, where $p=|H|$, because the Prüfer groups are the only indecomposable divisible torsion groups.

In particular $G$ is isomorphic to a subgroup of a Prüfer group; if it is finitely generated, it is cyclic; otherwise it coincides with the Prüfer group.

Answer 3

Since $G$ is finitly generatet abelian group;

$$G\cong \mathbb Z^n\oplus \mathbb Z_{n_1}\oplus...\mathbb Z_{n_k}$$

From the question; we have $G\cong \mathbb Z^n\oplus \mathbb Z_{n_i}$ where $n_i=p^n$.

since intersection of components trivial in the direct sum, $G\cong \mathbb Z$ or $G\cong \mathbb Z_{p^n}$. But in $\mathbb Z$, there is no integer which is divisible by all primes so f0r all $n\in Z$ there exist $pZ$ which does not contain it. Thus the answer is same $G\cong \mathbb Z_{p^n}$.