2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $G$. $\endgroup$ – Simon Dec 16 '16 at 8:55
  • 1
    $\begingroup$ Use $|HS|/|S| = |H|/|H \cap S|$. $\endgroup$ – Bungo Dec 16 '16 at 9:10
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Bungo's comment is by far the best answer, though. $\endgroup$ – Andreas Caranti Dec 16 '16 at 18:08
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Could you elaborate on how to get the contradiction? And why must it have index $p$? couldn't it be $p^2$, etc? $\endgroup$ – MathMajor Dec 16 '16 at 8:57
  • 1
    $\begingroup$ 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 $G$. $\endgroup$ – Rupert Dec 16 '16 at 9:08
  • $\begingroup$ Okay, I will try it on my own now. So I can check my answer, could you please edit yours with your intended solution? $\endgroup$ – MathMajor Dec 16 '16 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.