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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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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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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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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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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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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