# The intersection of a normal subgroup and Sylow $p$-subgroup

Let $$G$$ be a group and $$P\in Syl_p(G)$$, $$H$$ is normal in $$G$$. I want to show that $$P\cap H\in Syl_p(H)$$.

So I let $$P_0\in Syl_p(H)$$. $$P\cap H$$ is a $$p$$ subgroup of $$H$$, so by Sylow 2nd Theorem, $$P\cap H \leq P_0$$.

And by Sylow's 2nd and 3rd theorem, I get that there exists $$g\in G$$ such that $$P_0 \leq gPg^{-1}$$.

I think I want to prove that $$P_0 \leq P\cap H$$ next in order to conclude that $$P_0=P\cap H$$ but got stuck at this part.

You're almost there.

So, you have chosen a $P_0\in Syl_p(H)$ such that $P\cap H\le P_0$. Then, we have $P_0\le gPg^{-1}$. Also, $P_0\le H$, so, as $gHg^{-1}=H$, it means $$g^{-1}P_0\,g\le P\cap H\,.$$ Assuming everything is finite, by calculating sizes, we are ready, as $|g^{-1}P_0\,g|=|P_0|$ and both are included in $H$, so $|P\cap H|=|P_0|$ also follows.

• and this implies that $P_0=P \cap H$, right? – Akaichan Apr 3 '13 at 15:00

Hint: $PH$ is a subgroup when $H$ is a normal subgroup. Apply the formula

$$|PH| = \frac{|P||H|}{|P \cap H|}$$

• Is the subgroup generated by $P_{0}$ and $P$ a p-subgroup of G? I couldn't see your hint. Thanks! – Ergin Suer Jan 11 '14 at 21:04
• @Ergin Suer: First note that $PH$ is a subgroup of $G$ because $H$ is normal in $G$. The formula computes the index $(H:P \cap H)$ to $(H:P \cap H)= |H|/|P \cap H|=|PH|/|P|=(PH:P)$. Moreover, by the index formula, $(G:P)=(G:PH)\cdot (PH:P)$. Since $P$ is a Sylow $p$-subgroup of $G$, $(G:P)$ is coprime to $p$. Hence $(PH:P)$ is also coprime to $p$. Thus $(H:P\cap H)=(PH:P)$ is coprime to $p$. This shows that $P \cap H$ is a Sylow $p$-subgroup of $H$. – tj_ Jan 6 at 3:31

Suppose the converse, which is $$P \cap H$$ is not a Sylow $$p$$-subgroup of $$H$$. Then by Sylow theorems, there must exist a Sylow $$p$$-subgroup $$P'$$ of $$G$$ and $$q\in G$$, such that $$P'=qPq^{-1}\ \text{and}\ P\cap H Since $$H$$ is normal in $$G$$, we have: $$P'\cap H=(qPq^{-1})\cap H=q(P\cap H)q^{-1} but $$|P'\cap H|=|q(P'\cap H)q^{-1}|$$. Contradiction!
For non-normal subgroup $$H$$, the statement may not be valid. For example, take $$G=S_3$$ the symmetric group on three letters. Let $$H=\langle (12)\rangle$$ and $$P=\langle (13)\rangle$$ a Sylow $$2$$-subgroup of $$S_3$$, then clearly $$P\cap H=(1)$$ is not a Sylow $$2$$-subgroup of $$H$$.