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.