# Abstract Algebra: subgroups and order proof

• Nov 9th 2010, 04:55 PM
mulaosmanovicben
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?
• Nov 9th 2010, 05:59 PM
mulaosmanovicben
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
• Nov 9th 2010, 06:01 PM
tonio
Quote:

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
• Nov 9th 2010, 06:10 PM
mulaosmanovicben
so why can't G have order p^3 or greater?
• Nov 9th 2010, 06:17 PM
roninpro
Quote:

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 $\displaystyle 1, p, p^2, p^3$.
• Nov 9th 2010, 06:19 PM
mulaosmanovicben
why? that sounds like converse of lagranges theorem which is not always true.
• Nov 9th 2010, 08:07 PM
Drexel28
Quote:

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.
• Nov 10th 2010, 01:18 AM
tonio
Quote:

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
• Nov 10th 2010, 01:38 AM
Halmos Rules
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...