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Cyclic groups have at most one subgroup of any given finite index. Can we describe the class of all groups having such property?

Thank you!

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So, the natural question seems to be: 'Is it true that for any such group G, the canonical residually finite quotient of G (ie the quotient of G by the normal subgroup of elements that are contained in every finite-index subgroup) is cyclic?' Does anyone have a counterexample? –  HJRW Nov 27 '10 at 23:26
    
Actually, perhaps Arturo's answer proves exactly that. Arturo? –  HJRW Nov 27 '10 at 23:28
    
On further thought, I think so. I posted this as an answer. –  HJRW Nov 28 '10 at 5:32
    
OK, I now conjecture that arbitrary (ie possibly infinitely generated) $G$ has this property if and only if its profinite completion is a profinite cyclic group. I haven't had time to think about a proof, but I doubt it's hard. –  HJRW Dec 6 '10 at 6:11
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3 Answers

Since $G$ has exactly one subgroup of each finite index, and the index of a conjugate of $H$ equals the index of $H$, then every subgroup of finite index is normal.

If $G$ is finite, then every subgroup is normal, so the group must be a Dedekind group (also known as Hamiltonian groups).

All such groups that are nonabelian are of the form $G = Q_8 \times B \times D$, where $Q_8$ is the quaternion group of $8$ elements, $B$ is a direct sum of copies of the cyclic group of order $2$, and $D$ is an abelian group of odd order. Any of the factors may be missing.

Since $Q_8$ contains several subgroups of index $4$, if a factor of $Q_8$ appears then $G$ would have several subgroups of the same index, hence $G$ must in fact be an abelian group.

Since $G$ is finite and abelian, it is isomorphic to a direct sum of cyclic groups, $G = C_{a_1}\oplus\cdots\oplus C_{a_k}$, where $1\lt a_1|a_2|\cdots|a_k$. If $k\gt 1$, then $G$ contains at least two subgroups of order $a_{k-1}$; thus $k=1$ so $G$ is in fact cyclic. So the only finite groups with the desired property are the cyclic groups. If $G$ is infinite, you can have other possibilities. One example is the Prüfer group, Added: but only by vacuity: it has no proper subgroups of finite index.

In general, if $H$ if a subgroup of finite index in $G$ then $H$ is normal, as above, and $G/H$ also has the desired property and is finite; thus, $G/H$ is cyclic for every subgroup of finite index by the argument above. I'm sure there's more to be said, but I'll think about it a bit first...

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When you write "All such groups are of the form ..." you mean all such non-abelian groups. (An abelian group of any order is Dedekind, i.e. its 2-Sylow subgroup doesn't have to be elementary abelian.) –  Matt E Nov 12 '10 at 4:00
    
@Matt E: Yes; thank you. –  Arturo Magidin Nov 12 '10 at 4:02
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See Theorem 3.1 at math.uconn.edu/~kconrad/blurbs/grouptheory/sylowapp.pdf, which proves the result without using the classification of Dedekind groups. –  KCd Nov 12 '10 at 11:01
    
The link I gave before only treats the case of finite groups. –  KCd Nov 12 '10 at 11:02
    
@KCd: Thanks for the link. –  Arturo Magidin Nov 12 '10 at 15:27
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Just thought, I would make this obvious remark, extending Arturo's answer: since, even in the infinite case, any subgroup of finite index in $G$ must be normal for $G$ to satisfy the requirement, it follows that for any $H$ of finite index in $G$, any subgroup of the quotient $G/H$ will also correspond to a subgroup of $G$ of finite index. In particular, for any $H$ of finite index, $H$ must be normal and the quotient $G/H$ must be cyclic by Arturo's argument.

A class of such groups considerably extending that of cyclic groups are the pro-cyclic group, i.e. inverse limits of cyclic ones. Examples include $\mathbb{Z}_p$ or any product $\prod_p \mathbb{Z}_p$ over distinct primes $p$. In particular, $\hat{\mathbb{Z}}$ is another example. In fact, Arturo's argument shows that any pro-finite group satisfying the above condition must be pro-cyclic.

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rtel: I'm not that it is correct that such a group must be pro-cyclic. For instance, the Tarski monsters, en.wikipedia.org/wiki/Tarski_monster, satisfy the condition at hand vacuously (the only proper subgroups have infinite index), but are not pro-cyclic, since the only finite homomorphic images are trivial. –  Arturo Magidin Nov 26 '10 at 19:06
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The Tarski monster is clearly not pro-finite and is therefore not a counterexample to my observation. –  Alex B. Nov 27 '10 at 6:18
    
rtel: quite right; somehow, I completely skipped over the condition "pro-finite". Clearly, it's time to go get a new eyeglass prescription. Sorry about that. –  Arturo Magidin Nov 28 '10 at 5:38
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Let $G$ be a group. The canonical residually finite quotient of $G$ is $R(G)=G/K$ where $K$ is the intersection of all the finite-index subgroups of $G$.

Lemma: If $G$ is finitely generated (update) then $G$ has at most one subgroup of each index if and only if $R(G)$ is cyclic.

Proof: First, note that $R(G)$ is residually finite. If every finite quotient of $R(G)$ is cyclic then $R(G)$ is residually cyclic, and it follows that $R(G)$ is abelian. So $R(G)$ has a non-cyclic finite quotient unless $R(G)$ is cyclic. Therefore, if $R(G)$ is not cyclic then $R(G)$, and hence $G$, has a finite non-cyclic quotient, and hence, by Artuto's answer, has a two distinct finite-index subgroups of the same index.

Conversely, suppose that $R(G)$ is cyclic. Every finite-index subgroup of $G$ contains $K$, so the quotient map $G\to R(G)$ maps finite-index subgroups to finite-index subgroups bijectively and preserves the index. Therefore, if $R(G)$ is cyclic then $G$ has at most one subgroup of each index. QED

I believe that it is an open question whether or not there is an algorithm to determine whether a fp group has a proper finite-index subgroup, ie whether or not $R(G)$ is non-trivial. So it may be open whether or not it is possible to determine if $R(G)$ is cyclic, too.

Note: Earlier, I forgot to mention that I had implicitly assumed that $G$ is finitely generated. This assumption is clearly necessary; otherwise the additive group of the rationals is a counterexample. If $G$ is not finitely generated, then the same argument shows that if $G$ has at most one subgroup of each finite index then $R(G)$ is residually cyclic. But it's not clear to me that the converse of this statement is true. So I'll finish with a question:

If $G$ is residually cyclic, does $G$ have at most one subgroup of each finite index?

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Since interest in this thread seems to have died out, I've posted this as a question: math.stackexchange.com/questions/12635/… . –  HJRW Dec 1 '10 at 15:59
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