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

Code-Guru
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ROBINSON
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7 Answers7

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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.

Old John
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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.

Chris Leary
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  • 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
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    @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. – bzc Aug 22 '12 at 20:50
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    @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
  • Can we classify all the abelian torsion groups whose proper subgroups are cyclic? – Makoto Kato Aug 22 '12 at 21:01
  • @Makoto: such a group splits into its primary parts. Forcing every subgroup to be cyclic means either the group is cyclic, or is a p-group for some prime p. It then can be decomposed into its indecomposable parts, and insisting every subgroup is cyclic makes it either $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$, or the group in this answer, $\mathbb{Z}(p^\infty)$. –  Aug 22 '12 at 21:37
  • This group is also known as $\mathbf{Q}_p/\mathbf{Z}_p$. – Keenan Kidwell Aug 22 '12 at 23:42
  • Can we classify all the abelian groups whose proper subgroups are cyclic? – Makoto Kato Aug 22 '12 at 23:44
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    @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. –  Aug 23 '12 at 08:14
  • @SteveD Interesting. Thanks. – Makoto Kato Aug 24 '12 at 13:47
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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 $G$.

Seirios
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  • "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
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    @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 $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3 \simeq \Bbb Z_2 \times \Bbb Z_6$. – bzc 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
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    A point not made clear is that $\mathbb{Z}_p \times \mathbb{Z}_q, p,q$ prime is cyclic. – Ross Millikan Aug 22 '12 at 20:59
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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.

Matt E
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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.

Stefan Hansen
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user67117
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All such groups can be easily described:

They are isomorphic to either $C_p \times C_p$ for a given prime $p$, or to $C_{p^\infty}$, or to $Q_p$.

For finite groups it follows from the classification of finite abelian groups. For infinite abelian groups it is proved here

Chain Markov
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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.)

tom
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    Yes, but the OP specifically asked for a noncyclic abelian group all of whose proper subgroups are cyclic – Old John Aug 23 '12 at 00:01