# Thread: order of group G

1. ## order of group G

ord(G) = ( p^2)*(q^2)
( p , q = prime number)
G not has normal subgroup order p^2
I want to prove :

ord(G)=36

2. Do you know sylow's third theorem?

Also, do you know the theorem that states that every group of order $p^2$ is Abelian?

3. sorry, I don't know the answer!!
can you help me?

4. If you don't know Sylow's Third Theorem, then this problem is considerably harder.

However, trying to solve this problem, I can't help but feel that there might be some information missing from the problem.

Can you explain to me what you have covered in terms of group theory. What concepts/theorems have you learned thus far?

5. Originally Posted by Haven
If you don't know Sylow's Third Theorem, then this problem is considerably harder.

However, trying to solve this problem, I can't help but feel that there might be some information missing from the problem.

Can you explain to me what you have covered in terms of group theory. What concepts/theorems have you learned thus far?
The theorem does not hold. Use Sylow's theorem on a group of order $100=2^2 5^2$.

6. Originally Posted by Swlabr
The theorem does not hold. Use Sylow's theorem on a group of order $100=2^2 5^2$.

I think the OP meant: let G be a group of order $p^2q^2\,,\,\,p>q$ primes, and such that G has no normal subgroup of order $p^2$ ; then, it must be that $|G| = 36$.

We can argue as follows: any sbgp. of order $p^2$ in G is a Sylow p-sbgp. of G, so it is non-normal iff there's more than one such sbgps. Now, we know that the number $n_p$ of such sbgps. is congruent to $1\!\!\pmod p$ and that it divides $q^2$ , so we get that:

$n_p=q^2\Longrightarrow q^2=1\!\!\pmod p\Longrightarrow q=\pm 1\!\!\pmod p\Longrightarrow q=-1\!\!\pmod p$ (the other option is impossible: why?) , and as $p>q$ we must have that $p=q+1\Longrightarrow q=2\,,\,p=3$ and we're done.

Tonio

7. Oh - I missed that `not'.