# Thread: Abstract Algebra: subgroups and order proof

1. ## Abstract Algebra: subgroups and order proof

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?

2. If i can show G is cyclic this will be easy since for Each divisor of the order of a cyclic group there is a subgroup of that order

3. 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?

Suppose $\displaystyle p,q\mid |G|$ , with p,q two different primes, and then use Cauchy's theorem to show $\displaystyle G$ has at least 4 subgroups ...

After the above you already know G is a finite p-group, and now a simple counting argument finishes the task.

Tonio

4. so why can't G have order p^3 or greater?

5. Originally Posted by mulaosmanovicben
so why can't G have order p^3 or greater?
If your group has order $\displaystyle p^3$, then you are guaranteed to have subgroups of order $\displaystyle 1, p, p^2, p^3$.

6. why? that sounds like converse of lagranges theorem which is not always true.

7. Originally Posted by mulaosmanovicben
why? that sounds like converse of lagranges theorem which is not always true.
No, he is citing one of the Sylow Theorem's which says that a group has a subgroup of every prime multiplicity dividing it.

8. Originally Posted by mulaosmanovicben
why? that sounds like converse of lagranges theorem which is not always true.

With finite p-groups it is true...in fact, much more can be said: if $\displaystyle |G|=p^n$ , with p a prime, then $\displaystyle G$ has

a normal subgroup of order $\displaystyle p^k\,\,\forall\, 0\leq k\leq n$ .

Tonio

9. I think you can avoid Cauchy and Sylow and just use some basic results concerning cyclic groups. Here's a hint :
If {e}, H, G are the three distinct subgroups, what can you say about the subgroups of H? From this deduce the order of H and continue on from there...