# Thread: determine up to isomorphism all groups of order 13^2*3^2

1. ## determine up to isomorphism all groups of order 13^2*3^2

Here's my problem:

determine up to isomorphism all groups of order 13^2*3^2

so far I can do this

Let G be a group such that |G| = 13^2 * 3^2

n_3 = # of 3-sylow subgroups of G
n_13 = # of 13-sylow subgroups of G

n_3 = 1 (mod 3) and n_3 divides 13^2, thus n_3 = 1, 13 or 169
n_13 = 1 (mod 13) and n_13 divides 3^2, thus n_13 = 1

since the 13-sylow subgroup is unique, it must be normal in G

and here's where I go all fuzzy. Any help is appreciated.

I also found this thread: http://www.mathhelpforum.com/math-he...omorphism.html

which describes a similar problem, but wasn't sure if there would be further considerations since both primes are squared.

2. We have $|G| = 3^2 13^2 = 1521$.

Since the order of G has 2 distinct prime factors, 3 and 13, Sylow Theorem I (henceforth to be referred to as Sylow I) tells us that G has at least 1 Sylow-3 subgroup and at least one 1 Sylow-13 subgroup. Let's find out how many of each.

We will denote the number of Sylow-3 subgroups and Sylow-13 subgroups by $n_3 \text{ and } n_{13}$, respectively.

Sylow III tells us that the number of subgroups of G of order $3^2$ is congruent to 1 modulo 3 and is a factor of $\frac{|G|}{3^2} = 169$. Likewise, the number of subgroups of order $13^2$ is congruent to 1 modulo 13 and is a factor of $\frac{|G|}{13^2} = 9$. Moreover, since a subgroup of a prime power order is a Sylow-p subgroup (and thus is isomorphic to every other Sylow-p subgroup), we have

$n_3 \equiv 1 \pmod3 \text{ and } n_3 | 169
\implies n_3 = 1, 13, \text{ or } 169$

$n_{13} \equiv 1 \pmod3 \text{ and } n_{13} | 9
\implies n_{13} = 1$

There are 13 isomorphism types of finite groups of order 1521. They are:

C1521
C169 : C9
C39 x C39
C507 x C3
C117 x C13
C3 x (C169 : C3)
C3 x ((C13 x C13) : C3)
C3 x ((C13 x C13) : C3)
C13 x (C13 : C9)
(C13 x C13) : C9
(C13 x C13) : C9
(C13 : C3) x (C13 : C3)
C39 x (C13 : C3)