# Let G be a finite group, $H \triangleleft G$ be a normal subgroup and $S$ be a Sylow subgroup of $G$. Show that $H \cap S$ is a Sylow subgroup of $H$.

Let G be a finite group, $H \triangleleft G$ be a normal subgroup and $S$ be a Sylow subgroup of $G$. Show that $H \cap S$ is a Sylow subgroup of $H$.

I'm not too sure where to start here. Assuming that $S$ is a Sylow $p$-subgroup of order $p^n$, I would need to show that $H\cap S$ also has order $p^n$. Then I'm stuck.

• No. $H\cap S$ won't have order $p^n$ in general. You want to show it is Sylow in $H$, not in the whole $G$. – Simon Dec 16 '16 at 8:55
• Use $|HS|/|S| = |H|/|H \cap S|$. – Bungo Dec 16 '16 at 9:10

Let $P$ be a Sylow $p$-subgroup of $H$. Since $P$ is a $p$-subgroup of $G$, by one of the Sylow theorems, there is $x \in G$ such that $P \le S^{x}$, or $P^{x^{-1}} \le S$.

Now $Q = P^{x^{-1}}$ is contained in $H$, as $H$ is normal in $G$, and has the same of order of $P$, and thus it is another $p$-Sylow subgroup of $H$.

Now $S \cap H \ge Q$, and since $S \cap H$ is a $p$-group contained in $H$, and contains the Sylow $p$-subgroup $Q$ of $H$, we have $S \cap H = Q$, as required.

• Bungo's comment is by far the best answer, though. – Andreas Caranti Dec 16 '16 at 18:08

We will assume that every finite group has a $p$-Sylow subgroup for all primes $p$ dividing its order.

Suppose that $H \cap S$ is not a $p$-Sylow subgroup of $H$. Let $K$ be a $p$-Sylow subgroup of $H$ containing $H \cap S$ and suppose that the index of $H\cap S$ in $K$ is $p^{k}$ where $k$ is an integer such that $k>0$. Let $a$ be an element of $K\setminus (H \cap S)$, with the property that $a^{p^{m}} \in H \cap S$ but $a^{p^{i}} \notin H \cap S$ for all integers $i$ such that $0\leq i<m$. Here $m$ is a positive integer less than or equal to $k$. Let $K'$ be the subgroup $\langle H \cap S, \langle a \rangle \rangle$. This is a $p$-group which is a subgroup of $H$, with the property that the index of $H\cap S$ in $K'$ is greater than $1$ and less than or equal to $p^{m}$. Consider the group $S':=\langle S, \langle a \rangle \rangle$. This group contains $K'$. Let $T$ be a set of representatives of the left cosets of $K'$ in $H$, and let $T'$ be a set of representatives of the right cosets of $H\cap S$ in $S$. If $g\in S'$ then $g=s_{1}a^{i_{1}}s_{2}a^{i_{2}}\ldots s_{r}a^{i_{r}}$ where $s_{i} \in S, 0\leq a_{i}<p^{m}$ for all integers $i$ such that $1\leq i\leq r$. We can re-write this as $hs_{1}s_{2} \ldots s_{r}:=hs$ for some $h\in H$ since $H$ is normal in $G$, and this in turn is an element of $HT'=TK'T'$. It may be that $TK'T'$ is strictly larger than $S'$, but we can achieve equality by shrinking $T$. This shows that the cardinality of $K'T'$ divides that of $S'$, and this in turn is larger than the cardinality of $(H\cap S)T'=S$. But the cardinality of $K'T'$ is a power of $p$ and so a $p$-Sylow subgroup of $S'$ must have larger cardinality than that of $S$, but this contradicts the hypothesis that $S$ was a Sylow $p$-subgroup of $G$.

Sorry if the write-up could have been a bit more elegant.

• Could you elaborate on how to get the contradiction? And why must it have index $p$? couldn't it be $p^2$, etc? – MathMajor Dec 16 '16 at 8:57
• We can choose $K$ so that the index is $p$ because if $M$ is a Sylow $p$-subgroup of $H$ of order $p^{n}$ then we can find a subgroup of $M$ of order $p^{k}$ for each $k$ such that $0\leq k<n$. But that doesn't matter, the argument works just as well if you choose $K$ so that the index is any power of $p$. Pick an element $k \in K \setminus (H \cap S)$, of order $p^{k}$. Consider a Sylow p-subgroup the group $\langle S, \langle k \rangle \rangle$, and show it has order greater than $S$. That's a contradiction with $S$ being a $p$-Sylow subgroup of $G$. – Rupert Dec 16 '16 at 9:08
• Okay, I will try it on my own now. So I can check my answer, could you please edit yours with your intended solution? – MathMajor Dec 16 '16 at 9:20