# Classifying groups using Sylow theorems

## Homework Statement

Let p and q both be prime numbers and p > q. Classify groups of order p2q if p is not congruent to +1 or -1 mod q.

## The Attempt at a Solution

It is clear that the Sylow theorems would be the things to use here. So I guess this says that the number of subgroups of order q must be either 1, p, or p2 & congruent to 1 mod 1. And the number of subgroups of order p2 must be 1 or q and congruent to 1 mod p.
I think there has to be 1 normal subgroup of order p2 since only one satisfies those conditions.
For subgroups or order q it seems that there can be only one normal subgroup as well.

What I'm not sure about is how to go about find which groups are isomorphic to these groups of order p2q.

If H and K are normal in the group G, |H| |K| = |G|, and H intersect K is trivial, what is the structure of G?

Well, I think G has to be abelian. So my hunch is that G will be isomorphic to Zp2 x Zq or Zp x Zp x Zq.

Correct. What I was getting at before was that, in general, if H and K are normal in G, |H| |K| = |G|, and H intersects K trivially, then G ~ H x K (where ~ denotes isomorphism and x is a direct product). This is easy to prove. First show that HK = G, and then construct the most obvious homomorphism possible between H x K and G. The other conditions on H and K will show that this homomorphism is well-defined and injective. You should then be able to verify that the conditions of this lemma are met by Hp2 and Hq in G (where H denotes a Sylow subgroup).

However, you're not done yet. There's only one isomorphism type for groups of order q (i.e., cyclic). How may types are there for groups of order p2?

Edit: I think I misunderstood your notation. Does "Zr" mean the cyclic group of order r?

"Edit: I think I misunderstood your notation. Does "Zr" mean the cyclic group of order r? "
Yes that is what I meant with that notation.

"How may types are there for groups of order p2?"
I'm not sure. I guess the cyclic groups and direct product of cyclic groups?

I guess the cyclic groups and direct product of cyclic groups?
Right. To prove it, use the fact that all groups of order p2 are abelian. (If you didn't already know this, you can prove it using the class equation. This will get you as far as concluding that a group of order p2 has nontrivial center; for the rest, use the fact that if G/Z(G) is cyclic, then G is abelian, where Z(G) is the center of G.) Suppose |G| = p2 and G is not cyclic, and let x be a nonidentity element of G. What is the order of x? Consider < x > (the subgroup generated by x). Since G is abelian, you know that < x > is normal in G. You can produce another subgroup isomorphic to < x > that intersects < x > trivially (why?). Now use the lemma I gave you before to pin down the structure of G.