Some more p-Sylow subgroups and Normalizers

[barbutzo]

Homework Statement

Let G be a finite group, H a normal subgroup of G and B a p-Sylow subgroup of H for some p dividing |H|.

1. Let q be a prime dividing |G|, q != p, Q be a q-Sylow subgroup of G. Prove that Q acts on B by conjugation.
2. Let Q' be the image of Q in G/H, show that if |Q'|=|Q| then KxB is (isomorphic to) a subgroup of G, where K is a kernel of the action in (1).
3. Suppose Q is normal in G, does that imply that the image of KxB in G/H is normal in G/H?

Homework Equations

The standard theory of the Sylow theorems.
In the previous parts of the question morphism here helped me prove that G=HN_G(B) where N_G(B) is the normalizer of B in G, and that G/H is isomorphic to N_G(B)/N_H(B)

The Attempt at a Solution

(1) Actually, I think there must be some mistake in the question. For Q to act on B we need that Q be a subset of the normalizer of B in G, and that doesn't seem to be the case. It might be that I'm missing something here...

(2) This would imply that the intersection of Q and H is trivial, but I'm not sure how K would "look like" to gain intuition on how this could unfold.

(3) I think not, but if I could present G as some direct product that might give me a reason why.

 

[mighty2000]

Hello! First of all, it is important to note that the statement in part (1) is correct. Q does act on B by conjugation, even without the assumption that Q is a subset of the normalizer of B in G. Here is a proof:

Let b be an element of B and q be an element of Q. Then, since B is a p-Sylow subgroup of H, b is also a p-Sylow subgroup of H, and thus b is contained in some conjugate of B in H. Let h be an element of H such that b^h = b. Then, since H is a normal subgroup of G, q^h is also an element of Q. Therefore, we have q^h * b * (q^h)^(-1) = b^h = b, so Q acts on B by conjugation.

Moving on to part (2), let's first consider the case where K is trivial. In this case, Q acts on B by conjugation as a subgroup of the automorphism group of B. Therefore, the image of Q in G/H, denoted by Q', also acts on B by conjugation as a subgroup of the automorphism group of B. This implies that the map Q' x B -> B given by (g, b) -> g*b*g^(-1) is a group homomorphism, and since |Q'| = |Q|, this map is also injective. Thus, Q' is isomorphic to a subgroup of the automorphism group of B, and since B is a subgroup of G, this implies that Q' is isomorphic to a subgroup of G.

Now, let's consider the case where K is non-trivial. In this case, K acts on B by conjugation, and thus the image of K in G/H, denoted by K', also acts on B by conjugation. Similarly to the previous case, this implies that the map K' x B -> B given by (g, b) -> g*b*g^(-1) is a group homomorphism, and since |K'| = |K|, this map is also injective. Therefore, K' is isomorphic to a subgroup of the automorphism group of B, and since B is a subgroup of G, this implies that K' is isomorphic to a subgroup of G.

Finally, for part (3), it is