Are Normal Subgroups and p-Groups Key to Understanding Finite Groups?

[Poirot1]

Let G be a finite group and N a normal subgroup of G. Assume further that N is a p -group for some prime p.
1) By considering G/N, show that there is a subgroup H of G contaning N such that p does not divide [G].

2) Show that N is a subgroup of all p-subgroups of G.

My thoughts: for 1) maybe use sylow theorems. for 2) I don't understand because if N is a normal p-subgroup, then we have gNg^-1=N for all g. But by sylow, every p-subgroup is conjugate so we must conclude that N is the only p-subgroup of G.

 

[Deveno]

your reasoning for #2 is fine:

if N is a subgroup of a sylow p-subgroup P, then certainly gNg^-1 is a subgroup of gPg^-1.

you're on the right track for #1: let H be a sylow p-subgroup of G.

then |G| = |H|m, where p does not divide m, and [G] = |G|/|H| = |H|m/|H| = m.

 

[caffeinemachine]

Let G be a finite group and N a normal subgroup of G. Assume further that N is a p -group for some prime p.
1) By considering G/N, show that there is a subgroup H of G contaning N such that p does not divide [G].

2) Show that N is a subgroup of all p-subgroups of G.

My thoughts: for 1) maybe use sylow theorems. for 2) I don't understand because if N is a normal p-subgroup, then we have gNg^-1=N for all g. But by sylow, every p-subgroup is conjugate so we must conclude that N is the only p-subgroup of G.

I think use of sylow theorem in (1) helps.
Let $o(G)=p^nm,$ where $p$ doesn't divide $m$. Let $o(N)=p^r$. Let $K$ be a sylow p-subgroup of $G/N$. So $o(K)=p^{n-r}$. Then we know that (by what is probably called the fourth isomorphism theorem) $\exists H\leq G$ such that $N\leq H$ and $H/N \cong K$. It follows that $o(H)=p^n$. Thus we have shown that $N$ is contained in $H$ which is a sylow p-subgroup of $G$ and this is enough to prove (1).

 

[Poirot1]

what is the point of question 2) when N is the only p-subgroup? Wait, there's no difference between a sylow p-subgroup and a p-subgroup is there?

 

[caffeinemachine]

what is the point of question 2) when N is the only p-subgroup? Wait, there's no difference between a sylow p-subgroup and a p-subgroup is there?

Yeah I think there ain't any difference between a sylow p-subgroup and a p-subgroup. But p-subgroups and p-groups are different. So (2) ain't exactly trivial.
We are given a p-group N, which is a normal subgroup of G. We need to prove that it is a subgroup of all the p-subgroups.
Using (1) this is easy.

 

[Poirot1]

Yeah I think there ain't any difference between a sylow p-subgroup and a p-subgroup. But p-subgroups and p-groups are different. So (2) ain't exactly trivial.
We are given a p-group N, which is a normal subgroup of G. We need to prove that it is a subgroup of all the p-subgroups.
Using (1) this is easy.

What is the difference ? Is it that (sylow) p subgroups have order p^m where p^m is the highest power of p dividing o(G), whearas a p-group has order p^r where r is between 1 and m? That's the only way I can make sense of it.

 

[caffeinemachine]

What is the difference ? Is it that (sylow) p subgroups have order p^m where p^m is the highest power of p dividing o(G), whearas a p-group has order p^r where r is between 1 and m? That's the only way I can make sense of it.

This is the way I have made sense of it too.

 

[Deveno]

sylow p-subgroups are maximal p-subgroups.

 

[Poirot1]

by sylow theorems, if we let t represent the number of sylow p-subgroups of order say p^n, then there exist m,k such that $t-1=mp$ and $tk$=$\frac{|G|}{p^n}$

How do you know that the order of G is $mp^n$ when m does not divide p?

 

[Deveno]

by sylow theorems, if we let t represent the number of sylow p-subgroups of order say p^n, then there exist m,k such that $t-1=mp$ and $tk$=$\frac{|G|}{p^n}$

How do you know that the order of G is $mp^n$ when m does not divide p?

because EVERY positive integer is of that form (although n might be 0). although if we have a p-subgroup N, by lagrange n can't be 0.