# I Sylow subgroup of some factor group

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1. Mar 30, 2016

### moont14263

Hi. I have the following question:

Let $G$ be a finite group. Let $K$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. Let $P$ be a Sylow $p$-subgroup of $K$. Is $PN/N$ is a Sylow $p$-subgroup of $KN/N$?

Here is what I think.

Since $PN/N \cong P/(P \cap N)$, then $PN/N$ is a $p$-subgroup of $KN/N$.

Now $[KN/N:PN/N]=\frac{|KN|}{|N|} \frac{|N|}{|PN|}= \frac{|KN|}{|PN|}= \frac{|K||N|}{|K \cap N|} \frac{|P \cap N|}{|P||N|} = \frac{|K||P \cap N|}{|P||K \cap N|}=$ $[K:P]\frac{|P \cap N|}{|K \cap N|}$

Since $P$ is a Sylow $p$-subgroup of $K$, then $p$ does not divide $[K:P]$. Also, $p$ does not divide $\frac{|P \cap N|}{|K \cap N|}$ as $\frac{|P \cap N|}{|K \cap N|} \leq 1$ because $P \cap N$ is a subgroup of $K \cap N$. Therefore $p$ does not divide $[KN/N:PN/N]$.

Thus $PN/N$ is a Sylow $p$-subgroup of $KN/N$.

Am I right?

2. Apr 3, 2016

### andrewkirk

Looks broadly OK to me. I just have a few questions.
1. How do we know that $PN/N$ is a $p$-group?

2. This step seems to be doing a number of things, the validity of none of which is obvious to me. Can you justify it? I can get to the RHS from $[KN/N:PN/N]$, but not by the route shown above. I would use the 2nd isomorphism theorem and the fact that, for finite groups, $[A:B]= \frac{|A|}{|B|}$.

3. You can stop before the equals sign and just observe that, since $P$ is a Sylow $p$-subgroup of $K$, $p$ does not divide $\frac{|K|}{|P|}$, since the multiplicity of $p$ in the order of $|P|$ must equal the multiplicity of $p$ in the order of $|K|$.

3. Apr 5, 2016

4. Apr 5, 2016

### Staff: Mentor

You are free to do what you want. But this is a cross-post. Asking the same question at several different sites. This means Bungo and Andrew both worked out an answer. If you believe it is your right to do this, the net result of persistent cross-posting is a strong possibility that folks will skip your posts next time they see them.

I already know I will. Bye.