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How to find all groups that have exactly 3 subgroups?

Any group must have identity and itself as subgroups, so we just need to find all the groups that only have one proper subgroup. I think that for a prime $p$ the group $\mathbb Z/p^2\mathbb Z$ has only one proper subgroup (for example, $\mathbb Z/4\mathbb Z$, $\mathbb Z/9\mathbb Z$). Are there any other possibilities?

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1 Answer 1

Note that if two distinct primes divide the order of the group, then it will have subgroups of each of these orders, so only one prime can divide the order of the group. Also, a $p$-group has subgroups of any order dividing the order of the group, so the order must be $p^2$. Finally, a group is cyclic iff it has exactly one subgroup of any order dividing the order of the group, hence our group must be cyclic. In conclusion, $\mathbb{Z}/p^2\mathbb{Z}$ is the only possibility, and these satisfy the property for all primes $p$.

(For infinite groups, it is easy to check that this property can never hold.)

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There might be something wrong with my understanding. I don't get the "cyclic" part of the argument. Why ${\mathbb Z}\big/p^2{\mathbb Z}$ is the only possibility? ${\mathbb Z}_2\oplus{\mathbb Z}_2$ has $3$ subgroups up to isomorphism. –  Jack Sep 23 '13 at 22:03
    
@Jack No, it has $5$, when we include the entire group and the trivial subgroup (which we do here). We are not dealing with the subgroups up to isomorphism. –  Tobias Kildetoft Sep 24 '13 at 8:33

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