Cyclic group has 3 subgroups, what is the order of G

[Mr Davis 97]

Homework Statement

Suppose a cyclic group, G, has only three distinct subgroups: e, G itself,
and a subgroup of order 5. What is |G|? What if you replace 5 by p where
p is prime?

Homework Equations

The Attempt at a Solution

So, G has three distinct subgroups. By Lagrange's theorem, the order of the subgroup has to divide the order of the group. So the order of G is a multiple of 5. If we let |G| = 5, then there are only two subgroups, G and e. So we try |G| = 10. Why couldn't 10 be correct? Am I neglecting the fact that G is cyclic? (I know that the answer is actually 25, but am not sure why).

 

[Stephen Tashi]

Why couldn't 10 be correct?

The wording of the problem implies that G can't have a subgroup of order 2.

 

[Mr Davis 97]

The wording of the problem implies that G can't have a subgroup of order 2.

But if G has order 10 does that mean that it must necessarily have a subgroup of order 2 and a subgroup of order 5? I thought that lagranges theorem was just an if statement and not an if and only if statement.

 

[willem2]

For cyclic subgroups, Lagranges Theorem does work both ways. If k divides |G|, then {e, a^k, a^(2k), ... a^(|G| -k }, is a subgroup of G

 

[fresh_42]

A cyclic group $G$ is Abelian, so all subgroups $U$ are normal subgroups and thus a direct factor. This means $G/U$ is also (isomorphic to) a subgroup $V$ of $G$.

 

[pasmith]

But if G has order 10 does that mean that it must necessarily have a subgroup of order 2 and a subgroup of order 5? I thought that lagranges theorem was just an if statement and not an if and only if statement.

A finite group of even order must have an element of order 2: Partition the group into subsets \{a, a^{-1}\}. Such a set contains two elements if a^2 \neq e or one element if a^2 = e.

One of these subsets is \{e\} but the group order is even, so there must be at least one other subset consisting of a single element.