Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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

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 $2$ (exactly three, in fact), 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...

share|cite|improve this answer
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
See Theorem 3.1 at, 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

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.

share|cite|improve this answer
rtel: I'm not that it is correct that such a group must be pro-cyclic. For instance, the Tarski monsters,, 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
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

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?

share|cite|improve this answer
Since interest in this thread seems to have died out, I've posted this as a question:… . – HJRW Dec 1 '10 at 15:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.