I am trying to show that any group of order p2q has a normal Sylow subgroup where p and q are distinct primes.

In the case p>q I have no problem.. By Sylow np|q, so np is either 1 or q. But we also have n_p\equiv1\mod p which rules out q. So in the case p>q, n_p=1 and the unique p-Sylow subgroup is normal.

In the case p<q however, I run into a problem.. Again we have n_q|p^2 (so n_q\in{1,p,p^2}). Again, the condition n_q\equiv1\mod q rules out p.

Now I am attempting to rule out the case n_q=p^2: Assume n_q=p^2. Then p^2\equiv1\mod q\implies (p+1)(p-1)\equiv0\mod q\implies q must divide (p+1) since p<q and q prime. But since p<q and q|(p+1) we see q=p+1. For primes this only happens when p=2, q=3.Does this mean I need to check groups of order 2^2\cdot3=12 or did I miss something along the way that lets me conclude n_q\neq p^2 and thus that n_q=1.

**Answer**

Here's another approach: suppose n_q=p^2. The p^2 q-Sylows each have q-1 nonidentity elements, and since a group of prime order is generated by any of its nonidentity elements, these sets are disjoint. So our group has p^2(q-1) elements of order q, and so only p^2 elements of any other order. There's thus only room for one p-Sylow, which is thus normal.

This possibiity does in fact occur: A_4 has order 12, has the Klein group as a normal 2-Sylow, and all 8 other elements are 3-cycles.

**Attribution***Source : Link , Question Author : RHP , Answer Author : Chris Eagle*