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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.

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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
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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
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6 Answers 6

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.

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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.

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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. –  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
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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. –  user641 Aug 23 '12 at 8:14
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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 $G$.

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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
    
@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
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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.

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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.

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Did you mean Cayley table? –  Martin Sleziak Mar 19 '13 at 9:38
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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.)

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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 0:01
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