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Let $G=Z(p^k)$ where $k=1,2,..,\infty$. The group $G$ has exactly one subgroup series.

Does exist an other infinite abelian group with this property?

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You may find this fact generally useful: An abelian group that is not torsion-free has a direct summand isomorphic to $Z(p^k)$ for some $k=1,2,\ldots,\infty$. The answers below use a similar idea of having some sort of independent subgroup of this form, but the direct summand is particularly easy to use. – Jack Schmidt Sep 27 '12 at 13:06
up vote 2 down vote accepted

For any prime number $p$, the group of roots of unity whose order is a power of $p$ (the direct limit of cyclic groups of order a power of $p$) is an example of an infinite group with the mentioned property. (It is unclear to me which group is mentioned in the question, maybe it is this one?)

Since the property necessarily applies to all subgroups of a group $G$ that has it, and neither $\mathbf Z$ nor any group that non-trivally decomposes as a direct product has the property (so that $G$ is a torsion group, and the order of any element of $G$ must be a prime power), one easily shows that these examples are up to isomorphism the only ones.

Concretely: $\mathbf Z$ has uncomparable subgroups, for instance $2\mathbf Z$ and $3\mathbf Z$, while a finite cyclic group whose order has at least two different prime factors decomposes as a direct product by the (additive part of) the Chinese remainder theorem; neither of these groups has a unique chain of subgroups.

If $p$ is the unique prime factor of some nontrivial element of $G$, then the order of any element must be a power of $p$, lest we can find elements whose order has two distinct prime factors. Any two elements of order $p^k$ must generate the same subgroup of order $p^k$ of $G$ (otherwise those subgroups would be incomparable), and given that $G$ is infinite, it must have such a subgroup for all $k\in\mathbf N$. Then $G$ is the union of these subgroups.

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If $G$ is a finite abelian group, we can show that if the subgroup lattice is a chain, then $G$ is cyclic of prime power order, and hence the answer is no, at least in the finite case.

First, if $|G|$ is divisible by two distinct primes $p$ and $q$, then $G$ contains a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$, and neither of these is contained in the other (by Lagrange for example), so the subgroup lattice is not a chain.

So we can assume that $|G| = p^n$ for some prime $p$. If $G$ is not cyclic, then $n\geq 2$ and $G$ contains a subgroup isomorphic to $\mathbb{Z}/p\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$. If the subgroup lattice of $G$ is a chain, so is the subgroup lattice of any subgroup of $G$, so it is enough to show that the subgroup lattice of $\mathbb{Z}/p\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$ is not a chain.

But $\mathbb{Z}/p\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$ contains the two subgroups $\left<(1,0)\right>$ and $\left<(0,1)\right>$ (so here $1$ is the generator of $\mathbb{Z}/p\mathbb{Z}$ and $0$ is the identity). But neither of these subgroups is contained in the other, as their intersection is trivial.

This proves the original claim.

For the infinite case: If $G$ contains both an element of finite order and one of infinite order, then these generate subgroups neither of which is contained in the other, so we can rule out this.

If $G$ is torsion and contains elements of orders $p$ and $q$ for different primes $p$ and $q$, we can use the same argument as in the finite case.

If $G$ is torsion-free and not cyclic, then $G$ contains two elements of infinite order, neither of which is contained in the group generated by the other, so we do not get a chain as subgroup lattice.

We have now reduced this to loking at infinite $p$-groups. For any $n$, the elements of order dividing $p^n$ must be a cyclic subgroup, and since any element is contained in such a subgroup, we get that $G$ is indeed the direct limit of $\mathbb{Z}/p^k\mathbb{Z}$ for $k\geq 1$.

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thanks,for finite case it is trivial.I looking for infinite case. – Babak Miraftab Sep 27 '12 at 10:15

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