Let $P$ be a Sylow p-subgroup of a finite group $G$ and let $H$ be a normal subgroup of $G$. Show that if $p$ does not divide $[G:H]$, then $P \subseteq H$

I do not understand how to work this problem. If P is a Sylow p subgroup, then the order is prime or a power of a prime. We have to show that $P \subseteq H$, so does that mean that P is cyclic? But we must also show that $[G:H]$, so how do you prove $P \subseteq H$ but disprove $p$ does not divide $[G:H]$?? Any help in understanding would be appreciated.

After looking at the comments and taking some time to look at the problem, i am still somewhat confused. Here is a little of what I've understood.

If I let $|G|=p^an$ where $p \not| n$, then what I am trying to say is that I want to prove $p^an=|G|=|H|$ where $p$ in in $|H|$ and $p$ is not in $[G:H]$ which essentially means, that $|H|=p^an, p\not| n$ and $p \not | [G:H]$

Which I believe will show that $S$ will be a conjugate with $P$. $S$ is also a Sylow p-subgroup of $H$ Here is where I have stopped.


Observe that under the canonical homomorphism, and using the second (or third or something) isomorphism therem:

$$PH/H\cong P/(P\cap H)$$

It's clear the rightmost expression has order a power of $\;p\;$, yet the leftmost expression is clearly a subgroup pf $\;G/H\;$ , which by assumption is not divisible by $\;p\;$ .

From here that it must be that

$$\;PH/H=1\implies PH=H\implies P\le H$$

| cite | improve this answer | |
  • 1
    $\begingroup$ Or the same argument elementwise: If $g\in G$, then the order of its image $gH$ in $G/H$ divides $[G:H]$ and also divides the order of $g$. Hence for $g\in P$, $g$ has orde ra power of $p$ while $p\nmid [G:H]$, so the order of $gH$ is $1$, i.e., $g\in H$. $\endgroup$ – Hagen von Eitzen Mar 23 '15 at 6:33
  • $\begingroup$ Ok, I am still trying to understand the set up of this problem. I will add some comments to my original post $\endgroup$ – cele Mar 24 '15 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.