# $K$ is a normal subgroup of a finite group $G$ and $S$ is a Sylow $p$-subgroup of $G$. Prove that $K \cap S$ is a Sylow $p$-subgroup of $K$.

(This is my first post here, so please excuse me if I'm not following proper etiquette.)

First, I noted that both $K$ and $S$ are subgroups, thus their intersection is a subgroup. We can write that $|G| = p^km$ and $|S| = p^k$ where $p$ is a prime and $m$ has no factors of $p$. By Lagrange's theorem, since the intersection is a subgroup of S, then the cardinality of the intersection must be of the form $p^l$ where $l \leq k$. We can write $K$ as $p^rn$ where $r \leq k$ and $n$ has no factors of $p$. If I can show that $l=r$, then it seems that would be enough to show that the intersection is a sylow p-subgroup of K, but I can't seem to get anywhere else with anything.

I recall a theorem that states that since K is normal, then the intersection of K and S is normal in S but I don't know if that helps me much. There is the corollary of Sylow's 3rd theorem in my book that states that "A Sylow p-subgroup of a finite group G is a normal subgroup of G if and only if it is the only Sylow p-subgroup of G" which seems like I could use that somehow, but I'm not too sure.

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Let $M$ a Sylow $p$-subgroup of $K$. The Sylow theorems tell you something useful about $M$ and the Sylow $p$-subgroups of $G$. – Daniel Fischer Jan 16 '14 at 22:27
@DanielFischer I'm looking at the Sylow theorems and I don't see anything that tells me anything more about $M$ or the Sylow p-subgroups of $G$. I get that $M$ is conjugate to any other Sylow p-subgroup of $K$, don't know if that's useful. – benguin Jan 16 '14 at 23:01
It's a $p$-subgroup of $G$, so it is contained ine one of the conjugates of $S$, $M \subset gSg^{-1}$. Hence $g^{-1}Mg \subset S$. But also $g^{-1}Mg \subset K$ since $K$ is normal. – Daniel Fischer Jan 16 '14 at 23:05
How do we know that it's a p-subgroup of G? Suppose by example that $|G| = (2^3)(3), |K| = (2^2)(3), |M| = (2^2)$. $M$ is clearly a Sylow 2-subgroup of K but it is not a Sylow 2-subgroup of G. – benguin Jan 16 '14 at 23:21
It's in general not a Sylow $p$-subgroup of $G$. But its order is a power of $p$, so it's a $p$-subgroup of $G$. – Daniel Fischer Jan 16 '14 at 23:23

$K \cap S$ is a $p$-subgroup of $K$. Now $\frac{|KS|}{|S|} = \frac{|K|}{|K \cap S|}$, so $p \nmid [G : S] \implies p \nmid \frac{|KS|}{|S|} \implies p \nmid [K : K \cap S]$, i.e. $K \cap S$ is a Sylow-$p$ subgroup of $K$.
Ok, I'm going to go through your proof to see if I understand. So I get what you say about the second isomorphism theorem, I agree with that. Also, if p does not divide ($K:K \cap S$) that says something important. ($K: K \cap S) = |K|/|K \cap S|$. So if $p$ does not divide $|K|/|K \cap S|$, that must imply that the intersection has cardinality that how the same power of $p$ as $K$, which implies it is a Sylow p-subgroup. What I'm still unsure about is how result from the 2nd isomorphism theorem implies that p does not divide $(K : K \cap S)$. – benguin Jan 16 '14 at 23:15
I've edited the answer - the previous application of the second isomorphism theorem was incorrect, since $S$ need not be normal in $G$. The new argument only uses facts about orders – zcn Jan 16 '14 at 23:32
Let $N$ be a Sylow $p$-subgroup of $K$. $N$ is a $p$-subgroup of $G$ such that $N \leq xSx^{-1}$ for some $x\in G$. Then $N \leq xSx^{-1} \cap K=xSx^{-1} \cap xKx^{-1}$ since we have that $K$ is normal. Thus, $N \leq x(S \cap K)x^{-1}$, and consequently $x^{-1}Nx \leq (S \cap K)$. Now $x^{-1}Nx$ is a subgroup of $K$, and has maximal $p$-power order in $K$, since $x^{-1}Nx= \left|N\right|$. Since $S \cap N$ is a $p$-subgroup of $K$, $S \cap N$ is a maximal $p$-power subgroup of $K$, and hence a Sylow $p$-subgroup.