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.

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

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

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

  • $\begingroup$ Is the subgroup generated by $P_{0}$ and $P$ a p-subgroup of G? I couldn't see your hint. Thanks! $\endgroup$ – Ergin Suer Jan 11 '14 at 21:04
  • 1
    $\begingroup$ @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 $H$. $\endgroup$ – tj_ Jan 6 at 3:31

Let's prove by contradiction.

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 $$ P'=qPq^{-1}\ \text{and}\ P\cap H<P'\cap H .$$ Since $H$ is normal in $G$, we have: $$P'\cap H=(qPq^{-1})\cap H=q(P\cap H)q^{-1}<q(P'\cap H)q^{-1}$$ 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 $H$.


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