I'll be delighted to receive some guidance in the following questions:
1. Let G1,G2 be simple groups. Prove that every normal non-trivial subgroup of G= G1 x G2 is isomorphic to G1 or to G2...
2. Prove that every group of order p^2 * q where p,q are primes is solvable...
I've no idea about the second question. But the first one seems very easy, but I can't figure out how to solve it.
Help is needed
Thanks in advance
This is for 2.
Lemma 1. Any group of order is abelian for a prime number p.
This can be proven by using that any p-group has a non-trivial center (Use a class equation). If G/Z(G) is cyclic or {e}, then G is an abelian group.
Let and H be a Sylow q-subgroup of G. Then , which is abelian by Lemma 1. We remain to show that H is normal in G. Recall that the number of H (Sylow q-subgroup) in G is, denoted , divides |G| and having a form kq + 1. Now consider each case where q > p (If p>q, then you can show that there is a unique Sylow p-subgroup in G. This argument is relatively easy. If p=q, then we know that every finite p-group is solvable).
Case 1: .
H is normal in G, so .
Case 2: .
Since q > p, it never arises.
Case 3: .
In this case, there are Sylow q-subgroup of order q and there are q Sylow p-subgroup of order . If we have a unique Sylow p-subgroup of G, then we have a subnormal series, and we are done. Assume we have at least 2 Sylow p-subgroup of order in G. Any two distinct subgroups of order q intersect in the identity. Thus there are distinct elements of order q in G. Then, there should be distinct elements of order 1 or p or . By our assumption, we have at least 2 Sylow p-subgroup of order . Thus this situation neither arises by counting argument.
Thus and , which implies that G of order is solvable.