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

I've tried but I could not find a noncyclic Abelian group all of whose proper subgroups are cyclic. please help me.

share|cite|improve this question
How about $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$? Also known as the Klein four-group. – Old John Aug 22 '12 at 17:06
@OldJohn I think you can post this as an answer. – M Turgeon Aug 22 '12 at 17:12
As an extension to John's comment, I believe the only cases are $\mathbb Z/p\mathbb Z \oplus \mathbb Z/p\mathbb Z$ where $p$ is prime. – Thomas Andrews Aug 22 '12 at 17:14
@MTurgeon Is it then the correct etiquette on the site to delete my previous comment? (and this one.) – Old John Aug 22 '12 at 17:17
up vote 27 down vote accepted

The simplest possible example of this would be $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, as this is abelian and is the smallest group which is not cyclic. It is also known as the Klein four-group.

share|cite|improve this answer

The first example that came to mind, probably because I've spent so much time with it lately, is $\mathbb{Z}(p^{\infty})$, which is of course isomorphic to the group of all $p^n$-th roots of unity, $n=0,1,2, \ldots$. What I've always liked about this group is that all proper subgroups are finite as well as cyclic, while the group itself is infinite and non-cyclic. Plainly, the other examples are far simpler. Let this be a lesson to the OP: learn enough mathematics and you may easily overlook simple examples.

share|cite|improve this answer
A simpler presentation of this group is perhaps as the additive group of all rationals whose denominator is a power of $p$, i.e. $\{\ \frac{a}{p^i}\ |\ a \in \mathbb{Z},\, i \in \mathbb{N}\ \}$. – Peter LeFanu Lumsdaine Aug 22 '12 at 20:46
@Peter: I think you mean the subgroup of $\Bbb Q/ \Bbb Z$ with denominators a power of $p$. No subgroup of $(\Bbb Q, +)$ has torsion. – Brandon Carter Aug 22 '12 at 20:50
@Brandon: of course, yes — thankyou! Clearly I’m undercaffeinated today. – Peter LeFanu Lumsdaine Aug 22 '12 at 20:51
@Peter- I agree, but I couldn't decide which description would be easier for the OP. Maybe I'm undercaffeinated today, too! – Chris Leary Aug 22 '12 at 20:57
@MakotoKato: Yes, besides the cases I mentioned, there are the torsion free cases, which are always subgroups of $\mathbb{Q}$. There's some more advanced math here (basically you show every proper subgroup of $\mathbb{Q}$ is residually finite), but you eventually reduce to $\mathbb{Z}$, which is cyclic. In other words, there are only the two families I mentioned in my comment above. – user641 Aug 23 '12 at 8:14

More generally, any finitely generated noncyclic abelian group whose subgroups are cyclic has the form $\mathbb{Z}_p \times \mathbb{Z}_p$, where $p$ is prime.

Indeed, each finitely generated abelian group $G$ has the form $\mathbb{Z}_{n_1}\times ... \times \mathbb{Z}_{n_r} \times \mathbb{Z}^n$ with $n_1 \ | \ n_2 \ | \ ... \ | \ n_r$ and $n_1>1$.

Case 1: $n=0$. $G$ has to be noncyclic so $r\geq 2$. There is exists a prime $p$ dividing each $n_i$ and either $\mathbb{Z}_p \times \mathbb{Z}_p$ is a proper noncyclic subgroup of $G$ or $G= \mathbb{Z}_p \times \mathbb{Z}_p$.

Case 2: $n,r \neq 0$. $\mathbb{Z}_{n_1} \times ... \times \mathbb{Z}_{n_r} \times m \mathbb{Z}$ is a proper noncyclic subgroup of $G$.

Case 3: $r=0$ and $n \neq 0$. $G$ has to be noncyclic so $n\geq 2$. So $\mathbb{Z} \times m \mathbb{Z}$ is a proper noncyclic subgroup of $G$.

share|cite|improve this answer
"there is exists a prime $p$ dividing each $n_i$ (Chinese remainder theorem)." Could you please elaborate on this? – Makoto Kato Aug 22 '12 at 19:57
The $n_i$'s are not relatively prime so their greatest common divisor is not $1$. Then you can take a prime factor $p$ of the gcd. – Seirios Aug 22 '12 at 20:04
For example, how about $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$? Regards, – Makoto Kato Aug 22 '12 at 20:30
@Makoto: There is an additional condition inherent to the fundamental theorem of finitely generated abelian groups not mentioned above that guarantees uniqueness of this decomposition, namely $n_1 \mid n_2 \mid \dots \mid n_r$. So in your case $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3 \simeq \Bbb Z_2 \times \Bbb Z_6$. – Brandon Carter Aug 22 '12 at 20:42
I edited my answer using this argument. It is better, thank you. – Seirios Aug 22 '12 at 20:56

There is a general way to approach questions of the form "find a non-cyclic gp. (or ab. gp.) all of whose proper subgroups are cyclic", "find a non-ab. gp. all of whose proper sgs. are abelian", etc., namely:

Look for the smallest group that is non-cyclic/non-abelian/whatever.

Why does this work?

Well, if $G$ is non-cyclic, but any smaller group is cyclic, then any proper subgroup of $G$ will be cyclic. Since any group of order $< 4$ must be cyclic, we see that the Klein $4$-group (which is itself non-cyclic) satisfies the condition.

To check that you understand it, use this method to find a non-abelian group all of whose proper subgroups are abelian.

share|cite|improve this answer

Let us take a group $\{1,3,5,7\}$ with binary operation in $G$ as multiplication its modulu is $8$. If we draw a calas table we get to know that it is abelian but the elements of the groups do not seem to belong to a cyclic group.

share|cite|improve this answer
Did you mean Cayley table? – Martin Sleziak Mar 19 '13 at 9:38

One simple example would be the group of permutations of 3 elements. (This, as well as the other simple example already provided of the cross product of the 2-element group with itself, is a particular case of a more general solution: the cross product of two groups with a prime number of elements.)

share|cite|improve this answer
Yes, but the OP specifically asked for a noncyclic abelian group all of whose proper subgroups are cyclic – Old John Aug 23 '12 at 0:01

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.