Originally Posted by
mulaosmanovicben Hi sorry I did not know what to title this question!
Here is the problem:
Suppose that G has exactly three subgroups. Prove that G is finite and the order of G is a prime squared.
ATTEMPT:
G is either finite or infinite. Assume it is infinite. For g in G consider <g> ={g^n | n an integer} and <g^2> = {g^2n | n an integer} these both are subgroups but <g> =/= <g^2> so G has more than 3 subgroups so G cannot be infinite.
Now we must show that its order is a prime squared but i dont know what to do?
Any hints?