# Thread: Abstract Algebra - Normal groups

1. ## Abstract Algebra - Normal groups

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

2. Originally Posted by WannaBe
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

For the first one, consider that if $N\lhd G_1\times G_2$ then the set of first coordinates are invariant under conjugation, and so are the second...soo

3. Originally Posted by WannaBe
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

This is for 2.

Lemma 1. Any group of order $p^2$ 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 $|G|=p^2q$ and H be a Sylow q-subgroup of G. Then $|G/H| = p^2$, 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 $n_q$, 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: $n_q = 1$.
H is normal in G, so $\{e\} \triangleleft H \triangleleft G$.

Case 2: $n_q = p$.
Since q > p, it never arises.

Case 3: $n_q = p^2$.
In this case, there are $p^2$ Sylow q-subgroup of order q and there are q Sylow p-subgroup of order $p^2$. 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 $p^2$ in G. Any two distinct subgroups of order q intersect in the identity. Thus there are $p^2(q-1) = p^2q -p^2$ distinct elements of order q in G. Then, there should be $p^2$ distinct elements of order 1 or p or $p^2$. By our assumption, we have at least 2 Sylow p-subgroup of order $p^2$. Thus this situation neither arises by counting argument.

Thus $n_q = 1$ and $\{e\} \triangleleft H \triangleleft G$, which implies that G of order $p^2q$ is solvable.

4. Thanks a lot to both of you