Question

Prove that conversely any p-Sylow subgroup of H is also a p-Sylow subgroup of G if and only if |G : H| is not a μltiple of p.

a) True
b) False

Answer 1

The statement that any p-Sylow subgroup of H is a p-Sylow subgroup of G iff |G : H| is not a multiple of p is true, by reasoning through the divisibility properties of the indices and orders of the groups.

Step-by-step explanation:

The student asked for a proof concerning p-Sylow subgroups. The statement to prove is: "Conversely, any p-Sylow subgroup of H is also a p-Sylow subgroup of G if and only if |G : H| is not a multiple of p." This statement can be proven true.

First, let's denote a p-Sylow subgroup of H as P. If P is a p-Sylow subgroup of G also, it means that the order of P is p^n, where p^n is the highest power of p dividing |G|. Now, the order of H must be a product of a power of p (since P is a p-Sylow subgroup of H) and possibly other prime factors but not p itself, since the index |G : H| is not a multiple of p.

For P to be a p-Sylow subgroup of G, it is necessary that |G: P| is not divisible by p. We know that |G : P| = |G : H| * |H : P|. Since |H : P| is not divisible by p, if |G : H| is also not a multiple of p, then their product |G : P| will not be divisible by p. Thus, P is a p-Sylow subgroup of G. This aligns with the condition necessary for P being a p-Sylow subgroup of both H and G. Conversely, if |G : H| were a multiple of p, then |G : P| would be too, making P not a p-Sylow subgroup of G.