Number of Sylow $p$-subgroups is congruent to $1$ modulo $p^a$.

Question

Let $G$ be a finite group having more than one Sylow $p$-subgroup. Let $p^a$ be the least element of $\{|P|/|P\cap Q|: P,Q\in \operatorname{Syl}_p(G),P\neq Q\}$. Prove $n_p\equiv 1\pmod {p^a}$.

I do not know how to prove it. Could anyone give a help? Thanks.

Hint: Let $P$ be a Sylow $p$-subgroup $G.$ Let $P$ act by conjugation on all Sylow $p$-subgroups of $G$ and consider the length of the orbits other than ${P }$. — Geoff Robinson, Sep 04 '13 at 06:57
@JonathanY. I don't see how those two things can be unclear. There are only two natural choices for orderings, and they are equal in this case. And what else would the mod mean other than the usual? — Tobias Kildetoft, Sep 04 '13 at 07:00
Dear Prof. I am still confused could you explain more? thanks — Shiquan, Sep 11 '13 at 14:03

score 3 · Accepted Answer · edited May 22 '24 at 01:16

3

Before proving, I will state a lemma:

Lemma. If $P\in Syl_p(G)$ and $H$ is a $p$-group, then $H\cap{N_G(P)}=H\cap{P}$. (This is a standard lemma used for proving Sylow theorems.) (See bottom of page for proof.)

Now let $P$ be element of $Syl_p(G)$, and let $P$ act on $Syl_p(G)$ by conjugation. Denote $Q\in Syl_p(G)$ arbitrarily. Then every orbit $Orb_P(Q)$ has length $|P:P\cap{N_G(Q)}|$ by Orbit-Stabilizer Theorem. By our lemma, it is equal to $|P:P\cap{Q}|$. Notice that length of orbit is 1 if and only if $P=Q$. Now for rest of the orbits, find $R$ such that $|P:P\cap R|$ is as small as possible. Then $|P:P\cap R|$ divides the rest of the orbit lengths (which are $\neq1$). Since we know there is only one orbit with length one, it follows that $n_p≡1\ \text{mod} |P:P\cap R|$.

Proof of the Lemma.

Since $H\cap{N_G(P)}$ is a $p$-group in $N_G(P)$, then $H\cap{N_G(P)}$ must be contained in $P$. Thus, $H\cap{N_G(P)}\subset H\cap{P}$. The other inclusion is obvious, so we are done.

edited May 22 '24 at 01:16

J.G.131

1,128

answered Nov 12 '13 at 23:04

mesel

15,400

For anybody confused on the conclusion, note that $n_p=|Syl_p(G)|$. Since $|Syl_p(G)|=\sum_{Q\in Syl_p(G)}$$|Orb_P(Q)|$, and having shown that $|Orb_P(Q)|\equiv0\ \text{mod}(|P:$$P\cap R|)$ whenever $|Orb_P(Q)|\neq1$, we deduce $|Syl_p(G)|≡1\ \text{mod} (|P:P\cap R|)$. Fantastic response besides this little point of confusion! +1 – J.G.131 May 21 '24 at 23:57
The proof of that lemma might be circular. That lemma can be proved first to prove the Sylow theorems (for example see D&F Abstract Algebra page 140). But "Since $H\cap{N_G(P)}$ is a $p$-group in $N_G(P)$, then $H\cap{N_G(P)}$ must be contained in $P$." seems to assume some of the Sylow theorems. – 19021605 Sep 01 '25 at 13:03
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@19021605 :Let us say that $U=H\cap N_G(P)$. Then $U$ is a p group normalizing $P$, and so $UP$ is a p group containing $P$. Since $P$ is a Sylow, $UP=P$, and hence $U\subseteq P.$ – mesel Sep 02 '25 at 09:14

Number of Sylow $p$-subgroups is congruent to $1$ modulo $p^a$.

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