Sylow Theorem?

Question

Let N be a normal subgroup of a finite group G.Let p be a prime factor of order G s.t [G:N] is coprime to p
1)show N contains all Sylow p subgroups of G
2) Give an example to show 1) need not hold if N is not normal

Answer 1

The first part is rather easy (if I am thinking correctly).

First of all ([G,N],p)=1, thus p^a divides |N|, and there is at least one Sylow p-subgroup of N.
Let P be one of the sylow p-subgroups of N.
Now all the sylow p-subgroups are conjagate, so all of them can be written as xPx^(-1) for some x in G. But xPx^(-1) ⊆xNx^(-1)=N, so all of the conjugates of P (and thus all Sylow p-subgroups) are contained in N.

I need to go get some food, I'll think about the second part, shouldn't be that hard.


For the second part, just think of a Group with more than one Sylow p-group. For example think of the symmetric group over 4 elements (S4) and consider the Sylow 3-subgroups. Let N=<(1,2,3)> then |N|=3 and [G:N]=8 and (8,3)=1, so we have fulfilled the requirements of the statement (minus N being normal, of course). But <(1,2,4)> is clearly not in N, thus N does not contain all Sylow 3-subgroups.

Hope that all makes sense.

Answer 2

Dude, I don't think you'll find the answer here. Even if someone came up with the answer, I bet they'd have no clue what it means.