MHB Abstract Algebra Sylow Subgroup

[DavidL]

I have a question about abstract algebra so if someone could help me answering this question please ...

Suppose P,P' are 3-Sylow subgroup, and let Q be their intersection and N the normalizer of Q. Problem: Explain why is the order of N divisible by 9 ?

Thanks for your help.

Regards,

 

[Deveno]

Re: Abstrat Algebra Sylow Subgroup

Well, in general, that's not TRUE: for example, if $P$ is normal with $|P| = 3$ in the group $G$, then $P = P' = Q$ and $N_G(P) = G$, and $9 \not\mid |G|$ (or else $P$ would have larger order).

Are you leaving part of the problem out?

 

[DavidL]

Re: Abstrat Algebra Sylow Subgroup

Thanks for your answer but what is N_G(P)=G please ?

 

[Deveno]

Re: Abstrat Algebra Sylow Subgroup

$N_G(P)$ means the normalizer of $P$ in $G$. If $P$ is a normal subgroup, then all of $G$ normalizes $P$.

 

[DavidL]

Re: Abstrat Algebra Sylow Subgroup

Ok but in the case of the order of G is 180.
Suppose that P,P' are 3 Sylow subgroup and let Q be there intersection and N the normalizer of Q.

Explain why is the order of N divisible by 9 ?Thanks :-)

 

[Deveno]

Re: Abstrat Algebra Sylow Subgroup

That's a different story, now we have some more information to go on.

First, we factor 180 into primes:

$180 = 2^2\cdot 3^2\cdot 5$

This tells us that the sylow 3-subgroups have order 9.

You're probably trying to show that $G$ has a nontrivial proper normal subgroup (that is: that $G$ is not simple), so let's assume the sylow 3-subgroups are not normal in $G$.

Now, here, we can use a "trick": any group of order 9 is abelian, which means that $Q$ is normal in $P$, which means in particular, that $P$ normalizes $Q$ so that $P \subseteq N(Q)$.

Hence, by Lagrange, $9 = |P|$ divides $|N(Q)|$.