# If H is a subgroup of G, then H has no more Sylow subgroups than G

If $H$ is a subgroup of the finite group $G$, then how do I show that $n_p(H) \leq n_p(G)$? Here $n_p(X)$ is the number of Sylow $p$-subgroups in the finite group $X$.

Here is my attempt: Suppose the order of $G$ is $n$, the order of $H$ is $m$, then $m$ divides $n$ by Lagrange's theorem. By Sylow's counting theorem $n_p(H) = 1 \pmod p$ and $n_p(H)$ divides $m$, $n_p(G) =1 \pmod p$, $n_p(G)$ divides $n$.

How do I continue after this?

• It is unclear which facts you are given and what you are trying to prove. Is $p$ a prime? What do you mean by $p(G)$? If you clarify your question you are more likely to get some help solving it. – Mark Bennet Feb 28 '14 at 17:11
• yes, p is a prime, and np(g) means the order of sylp(G) – user132177 Feb 28 '14 at 17:17
• What does sylp mean? Are you trying to prove that if $H$ is a subgroup of $G$ then the order of $H$ is less than or equal to the order of $G$? – user1729 Feb 28 '14 at 17:20
• sylp(G):={sylow p-subgroups of G}, I want to prove if H is a subgroup of G then the order of sylp(H) is less than or equal the order sylp(G) – user132177 Feb 28 '14 at 17:24
• Every Sylow $p$-subgroup $Q$ of $H$ is contained in some Sylow $p$-subgroup $P$ of $G$. You need to prove that you cannot have two distinct Sylow subgroups $Q_1$ and $Q_2$ contained in the same Sylow $p$-subgroup $P$ of $G$. To prove that impossible, consider $P \cap H$ would contain both $Q_1$ and $Q_2$. – Derek Holt Feb 28 '14 at 17:29

I like this question, and wanted it to have a bit of a longer answer:

## Surprising

This result should be a little surprising. After all the Sylow $p$-subgroups of $H$ can have smaller order, and even though $G$ may have only a few subgroups of order $p^n$, it might have a ton of order $p^{n-1}$. For example, in $G=A_4$, we have only $n_2(G)=1$ Sylow $2$-subgroup, but it has $3$ subgroups of order $2^1$, and so a subgroup $H$ has $n_2(H) \leq 3$, but that leaves open the possibility of $n_2(H) \in \{2,3\}$ both of which contradict the theorem. There is an analogue of $A_4$ called $G=\operatorname{AGL}(1,p^2)$ with approximately the same behavior: $n_p(G)=1$ but $G$ has $p+1$ subgroups of order $p$. When one actually looks for a subgroup $H$, one runs into a problem: no single $H$ contains all those subgroups of order $p$ (or even two of them in the $A_4$ and AGL cases) unless it contains an entire Sylow $p$-subgroup (making those smaller $p$-subgroups irrelevant).

## Proof

This idea leads to a simple proof (given by Derek Holt in the comments).

We construction a 1-1 function $f$ from $\operatorname{Syl}_p(H)$ to $\operatorname{Syl}_p(G)$ where $\operatorname{Syl}_p(X)$ is the set of Sylow $p$-subgroups of the finite group $X$. Given a Sylow $p$-subgroup $Q$ of $H$, Sylow's containment theorem says that $Q$ (a $p$-subgroup of $G$) is contained in some Sylow $p$-subgroup $f(Q)$ of $G$. If $f(Q_1) = f(Q_2)$, then $Q_1, Q_2 \leq f(Q_1)$ and $Q_1, Q_2 \leq H$, so $Q_1, Q_2 \leq f(Q_1) \cap H$. However, $f(Q_1) \cap H$ is a $p$-subgroup of $H$, and a $p$-subgroup of $H$ can only contain at most a single Sylow $p$-subgroup of $H$, so $Q_1 = Q_2$. Hence $f$ is a 1-1 function, and $n_p(H) \leq n_p(G)$.

## Normal subgroups

If $H$ is a normal subgroup of $G$, then in fact $n_p(H)$ divides $n_p(G)$. Hall (1967) calculated $n_p(G) = n_p(H) \cdot n_p(G/H) \cdot n_p(T)$ where $T=N_{PH}(P \cap H)$ for any Sylow $p$-subgroup $P$ of $G$.

• Hall's result is really nice. – mesel Feb 28 '14 at 19:19
• Just to be clear, $T=N_{PH}(P \cap H)$ is perfectly fine. The quotient seems pretty silly to me. – Jack Schmidt Feb 28 '14 at 20:38
• Ah, you are right. For some weird reason Hall states his theorem with $N_{PH}(P \cap H) / P \cap H$. – Mikko Korhonen Feb 28 '14 at 20:57
• Yeah, I'm not sure why. I've been trying to put restrictions on $T$ based on $G$, but I don't think any of them would care about normal subgroups of $O_p(T)$. – Jack Schmidt Feb 28 '14 at 21:26
• Like imagine $G=S_n \wr S_m$, then $G$ doesn't have too many normal subgroups (certainly we know which simple groups have any chance of occurring in a composition series of $G$), but how bad can $T$ be? $G$ itself contains a bazillion simple groups, but only $A_n$, $A_m$, and $C_2$ are allowed as composition factors of $H$ or $G/H$. Can $T$ have others? – Jack Schmidt Feb 28 '14 at 21:28

If $Q \in \operatorname{Syl}_p(H)$, then $Q \leq P$ for some $P \in \operatorname{Syl}_p(G)$. Then $Q \leq P \cap H \leq H$ and $P \cap H$ is a $p$-group, so it follows that $Q = P \cap H$.

In other words, we have the inclusion $\operatorname{Syl}_p(H) \subseteq \{P \cap H: P \in \operatorname{Syl}_p(G) \}$ which gives the result. This inclusion is an equality when $H$ is a normal subgroup, but not in general.

This is really a consequence of the pigeonhole principle: Assume that $H$ has more Sylow $p$-subgroups than $G$. Since every Sylow $p$-subgroup of $H$ is contained in a Sylow $p$-subgroup of $G$, there exists a Sylow $p$-subgroup $P$ of $G$ containing two distinct Sylow $p$-subgroups $Q_1$ and $Q_2$ of $H$. Since $Q_1$ and $Q_2$ are distinct, the group generated by them is not a $p$-group; on the other hand, $<Q_1,Q_2>\le P$ which is a $p$-group, a contradiction.

• Why is it true that every Sylow-$p$ subgroup of $H$ is contained in a Sylow-$p$ subgroup of $G$? It's true that such subgroups must be $p$-groups, but I can't see why every $p$-subgroup would need to be contained in a Sylow $p$-subgroup. – Eric Stucky Feb 28 '14 at 19:42
• @EricStucky: This is by the definition of Sylow $p$-subgroups: A Sylow $p$-subgroup of $G$ is a maximal (w.r.t. inclusion, not order) $p$-subgroup of $G$. See en.wikipedia.org/wiki/Sylow_theorems. – jpvee Feb 28 '14 at 19:47
• Thank you. That was never how they were defined in my classes. But it's great that the second theorem still works in this context! – Eric Stucky Feb 28 '14 at 19:55
• Why do you conclude that the group generated by distinct Sylow $p$-subgroups of $H$ is not a $p$-group? – Bach Sep 25 '19 at 17:14
• Like I wrote in my answer to Eric above: Since $Q_1$ and $Q_2$ are distinct maximal $p$-subgroups of $H$, the group they generate can not be a $p$-group. – jpvee Sep 27 '19 at 10:12

Bonus information: there is a paper of prof Gabriel Navarro from 2003 that implies that for a solvable group $$G$$, and any prime $$p$$ dividing $$|G|$$, $$n_p(H) \mid n_p(G)$$ for every subgroup $$H$$ of $$G$$. Here $$n_p(X)=\#Syl_p(X)$$ (which is understood to be $$1$$ if $$p \nmid |X|$$).