Why Does G Equal N_G(P)N in Group Theory?

[bham10246]

This shouldn't be too hard but I'm having a hard time putting a few pieces together.

Let $G$ be a finite group with $N$ a normal subgroup. Let $P$ be a Sylow $p$-subgroup of $N$. Prove that $G=N_G(P)N$, where $N_G(P)$ denotes the normalizer of $P$ in $G$.

My attempts: I know that the number of conjugates of P in G equals the index of the normalizer of P in G. What I don't understand is: why does the number of conjugates of P in G equal the normal subgroup N?

 

[mathwonk]

well it looks plausible ebcause NP(G) is th stuff in G that leaves P inside P, and N moves P around to all its conjugates inside N. Since N is normal also all elments of G leave P inside of N, so if you have the stuff that leaves P fixed when acting by conjugation, and also have a transitive subgroup acting by conjugation, you should have everyone.

i.e. let G act on P by conjugation and prove NP(G)N contains a a transitive subgroup for this action, plus the full isotropy subgroup. then it is done isn't it? by the usual trick for finding automorphism groups. i.e. any group that acts transitively and contains an isotropy group of on point is the full automorphism group. (see cartan's wonderful little book on complex variables, in the section on complex automorphisms.)

 

[tacman]

The normalizer of a subgroup P in a group G is defined as the set of elements in G that commute with every element in P. In other words, the normalizer is the largest subgroup of G in which P is a normal subgroup. In this case, we have a finite group G with a normal subgroup N, and we are looking at the Sylow p-subgroup P of N.

First, we need to understand that the number of conjugates of P in G is equal to the index of the normalizer of P in G. This is because the number of conjugates of P in G is the number of distinct subgroups that are isomorphic to P, and each of these subgroups has the same index in G as the normalizer of P.

Now, to prove that G=N_G(P)N, we need to show that every element in G can be written as a product of an element in N_G(P) and an element in N. Let g be an element in G, we need to show that there exists an element n in N and an element n_p in N_G(P) such that g=n_pn.

Since P is a Sylow p-subgroup of N, we know that P is a subgroup of N with index p^k, where p is a prime number and k is a positive integer. This means that N has p^k subgroups that are isomorphic to P. Since the number of conjugates of P in G is also p^k, we can say that each of these subgroups is conjugate to each other.

Now, let n be an element in N, and let n_p be an element in N_G(P) such that n_p*P*n_p^-1=n. This means that n_p is an element in G that normalizes P, and thus, n_p is also an element in N_G(P). Therefore, we can write g=n_pn, where n_p is in N_G(P) and n is in N. This shows that every element in G can be written as a product of an element in N_G(P) and an element in N, and hence, G=N_G(P)N.

In conclusion, the number of conjugates of P in G is equal to the index of the normalizer of P in G, and this helps us prove that G=N_G(P)N. The normalizer plays a crucial role in