# Thread: Show P is normal in G

1. ## Show P is normal in G

Problem:

Let $\displaystyle P$ be a Sylow p-subgroup of $\displaystyle G$ and assume that $\displaystyle P \triangleleft N \triangleleft G$. Show that $\displaystyle P \triangleleft G$.

Could this be shown without giving out the cardinality of G first? I mean, for example without assuming first that G has a cardinality a power of a prime, or G has even cardinality, or anything like that. How do you start solving this?

2. Originally Posted by guildmage
Problem:

Let $\displaystyle P$ be a Sylow p-subgroup of $\displaystyle G$ and assume that $\displaystyle P \triangleleft N \triangleleft G$. Show that $\displaystyle P \triangleleft G$.

Could this be shown without giving out the cardinality of G first? I mean, for example without assuming first that G has a cardinality a power of a prime, or G has even cardinality, or anything like that. How do you start solving this?
Firstly, you know that P must be a Sylow p-subgroup og G (why?). You also know that conjugation by elements of G permutes through the Sylow p-subgroups of G. However, conjugation of P by elements of G will always keep you inside N. This means that conjugation of P by an element of G will give you a Sylow p-subgroup of N. Now, as P is normal in N it is unique and so P is a unique sylow p-subgroup of G.

3. Tell me if I got you right:

Well, $\displaystyle P$ is given to be a Sylow p-subgroup of $\displaystyle G$. I also know that each conjugate of $\displaystyle P$ is also a Sylow p-subgroup. Now $\displaystyle gP{g^{ - 1}} \subseteq N$ for all $\displaystyle g \in G$ because $\displaystyle N \triangleleft G$. Since $\displaystyle P \triangleleft N$, then $\displaystyle P$ must be the only Sylow p-subgroup of $\displaystyle N$. Since $\displaystyle P$ is a Sylow p-subgroup of $\displaystyle G$ (which is also the only Sylow p-subgroup of $\displaystyle G$), then $\displaystyle P$ must also be the only Sylow p-subgroup of $\displaystyle G$. Therefore, $\displaystyle P \triangleleft G$.

With this proof, however, I used the fact that if $\displaystyle P \triangleleft N$ then $\displaystyle P$ is unique. But the notes I have only guarantees the other way around: that if $\displaystyle P$ is the only Sylow p-subgroup of $\displaystyle N$, then $\displaystyle P \triangleleft N$. Is this statement a really biconditional? Because if it is, I would have to show it first.

4. Originally Posted by guildmage
Tell me if I got you right:

Well, $\displaystyle P$ is given to be a Sylow p-subgroup of $\displaystyle G$. I also know that each conjugate of $\displaystyle P$ is also a Sylow p-subgroup. Now $\displaystyle gP{g^{ - 1}} \subseteq N$ for all $\displaystyle g \in G$ because $\displaystyle N \triangleleft G$. Since $\displaystyle P \triangleleft N$, then $\displaystyle P$ must be the only Sylow p-subgroup of $\displaystyle N$. Since $\displaystyle P$ is a Sylow p-subgroup of $\displaystyle G$ (which is also the only Sylow p-subgroup of $\displaystyle G$), then $\displaystyle P$ must also be the only Sylow p-subgroup of $\displaystyle G$. Therefore, $\displaystyle P \triangleleft G$.

With this proof, however, I used the fact that if $\displaystyle P \triangleleft N$ then $\displaystyle P$ is unique. But the notes I have only guarantees the other way around: that if $\displaystyle P$ is the only Sylow p-subgroup of $\displaystyle N$, then $\displaystyle P \triangleleft N$. Is this statement a really biconditional? Because if it is, I would have to show it first.
Yes, it is a biconditional. The theorem in question reads as "All the Sylow p-subgroups are conjugate in G". The result you have been given is merely a corollary of this.