abstract-algebrafinite-groupsgroup-theorysylow-theory
We've been studying Sylow $p$-subgroups, and I've come across this problem.
Let $H$ be a subgroup of $G$, and suppose $G$ is finite. Then, $n_p (H) \leq n_p (G)$, where $n_p$ denotes the number of Sylow $p$-subgroups of a group.
I am having trouble figuring this one out, and I was wondering if anyone could help?
Thank you.
I like this question, and wanted it to have a bit of a longer answer:
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).
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 $n_p(H) \leq n_p(G)$.
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 $G$.
(See also this answer of Mikko Korhonen.)
Let $G$ be a group and let $$|G|=p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$$ where $p_1<p_2<\cdots<p_n$ are distinct primes and $a_i\geq 1$ for all $i$. Let $N_p$ denote the number of Sylow $p$-subgroups and let $N$ denote the total number of Sylow subgroups, i.e. the sum of all $N_{p_i}$. By the fact that $N_p\equiv 1\pmod{p}$ and $N_p||G|$ we must have that $$N_{p_i}\leq \frac{|G|}{p_i^{a_i}}$$
Suppose $p_i\neq 2$ and $a_i=1$. Then $N_{p_i}$ is equal to the number of elements of order $p_i$ divided by $p_i-1$. Thus if $i_1<i_2<\cdots<i_m$ satisfy $a_{i_k}=1$ for all $k$ and these are all the indices for which the exponent is $1$, then $$\sum_{k=1}^{m}{N_{p_{i_k}}}\leq \frac{|G|}{p_{i_1}-1}$$
New, more elementary proof
There are five cases.
(1) $a_i\geq 2$ for all $i$.
In this case $$N\leq\sum_{p}{\frac{1}{p^2}|G|}$$ where $p$ ranges over all primes, and the sum of the squares of the recpirocals of all primes is less than or equal to $\frac{453}{1000}$, so $$N\leq \frac{453}{1000}|G|<\frac{2}{3}|G|$$
(2) $p_1=2$, $G$ has a normal $2$-complement, and $N_2<\frac{1}{2}|G|$
Let $H$ be the normal $2$-complement. Then $H$ contains all Sylow subgroups for primes other than $2$. Suppose $H$ is of order $2^{m-a_1}|G|$, where $m<a_1$. By induction on $n$ we have that $$\sum_{k>1}{N_{p_k}}\leq \frac{2}{3}|H|=\frac{2^{m+1}}{3\cdot 2^{a_1}}|G|\leq\frac{1}{3}|G|$$ Since $N_2<\frac{|G|}{2}$ we have that $N_2\leq \frac{|G|}{4}$, hence $$N\leq N_2+\frac{1}{3}|G|\leq \frac{1}{4}|G|+\frac{1}{3}|G|=\frac{7}{12}|G|<\frac{2}{3}|G|$$
(3) $p_1=3$ or $p_2=3$ and $G$ has a normal $3$-complement
This is similar to the previous case.
Let $H$ be the normal $3$-complement. Then $H$ contains all Sylow subgroups for primes other than $3$. Suppose $H$ is of order $3^{m-a_1}|G|$, where $m<a_1$. By induction on $n$ we have that $$\sum_{k\neq 2}{N_{p_k}}\leq \frac{2}{3}|H|=\frac{2\cdot 3^{m-a_1}}{3\cdot 3^{a_1}}|G|\leq\frac{2}{9}|G|$$ So $$N\leq N_3+\frac{2}{9}|G|\leq \frac{1}{3}|G|+\frac{2}{9}|G|=\frac{5}{9}|G|<\frac{2}{3}|G|$$
(4) $G$ has no normal $3$ complement, and if $p_1=2$ then $G$ also has no normal $2$-complement
The condition implies that $a_1\geq 2$, unless $p_1\geq 5$. In this case neither the Sylow $2$-subgroup nor the Sylow $3$-subgroup is in the center of its normalizer (unless each respective subgroup is trivial), so by Burnside's transfer theorem we have $N_2\leq\frac{1}{8}|G|$ (which also holds if there is no nontrivial Sylow $2$-subgroup) and $N_3\leq\frac{1}{6}|G|$ (which also holds if there is no nontrivial $3$-subgroup). The contribution of the $N_{p_i}$ for $p_i\geq 5$ such that $a_i=1$ is as usual at most $\frac{1}{4}|G|$. Thus $$N\leq\sum{\frac{1}{p^2}|G|}-\frac{1}{4}|G|-\frac{1}{9}|G|+\frac{1}{8}|G|+\frac{1}{6}|G|+\frac{1}{4}|G|\leq \frac{2851}{4500}|G|<\frac{2}{3}|G|$$
(5) $p_1=2$ and $N_2=\frac{1}{2}|G|$
In this case as below we are in the situation of Suppose that half of the elements of G have order 2 and the other half form a subgroup H of order n. Prove that H is an abelian subgroup of G., meaning that we have an abelian subgroup of index $2$, hence every Sylow $p_i$ subgroup for $i>1$ is normal, hence $$N=\frac{1}{2}|G|+n-1$$ and since in particular $n-1\leq\frac{|G|}{6}$ because $|G|\geq 2\cdot 3^{n-1}$ we have that $$N\leq \frac{2}{3}|G|$$ This bound is tight, with $S_3$ giving an example of equality (in fact $S_3$ is unique in this sense, which can be seen from the proof).
Old, overpowered proof (no need to read, preserved for posterity)
Our proof is broken into six cases (case 4 has some subcases though):
(1) $a_i\geq 2$ for all $i$
In this case $$N\leq\sum_{p}{\frac{1}{p^2}|G|}$$ where $p$ ranges over all primes, and the sum of the squares of the recpirocals of all primes is less than or equal to $\frac{23}{50}$, so $$N\leq \frac{23}{50}|G|<\frac{2}{3}|G|$$
(2) $a_i=1$ for some $i$ and either $p_1\neq 2$ or $p_1=2$ and $a_1\geq 4$
For $p$ prime we have $$\sum_{p>3}{\frac{1}{p^2}}\leq \frac{23}{50}-\frac{1}{4}-\frac{1}{9}=\frac{89}{900}$$ If $p_1\geq 3$ we have $$N\leq \sum_{k=1}^{m}{N_{p_{i_k}}}+\frac{89}{900}|G|\leq\frac{1}{p_1-1}|G|+\frac{89}{900}|G|\leq \frac{1}{2}|G|+\frac{89}{900}|G|=\frac{539}{900}|G|<\frac{2}{3}|G|$$ If $p_1=2$ we have assumed $a_1\geq 4$. If $p_2>3$, or $p_2=3$ and $a_2=1$ then since $$\frac{1}{16}+\sum_{p>3}{\frac{1}{p^2}}\leq \frac{23}{50}-\frac{1}{9}-\frac{3}{16}=\frac{581}{3600}$$ we have $$N\leq \sum_{k=1}^{m}{N_{p_{i_k}}}+\frac{581}{3600}|G|\leq \frac{1}{p_2-1}|G|+\frac{581}{3600}|G|\leq\frac{2381}{3600}|G|<\frac{2}{3}$$ If $p_2=3$ and $a_2>1$, let $p_k$ be the smallest integer such that $a_k=1$. Then since $$\frac{1}{16}+\sum_{p\geq 3}{\frac{1}{p^2}}\leq \frac{23}{50}-\frac{3}{16}=\frac{109}{400}$$ we have $$N\leq \sum_{k=1}^{m}{N_{p_{i_k}}}+\frac{109}{400}|G|\leq \frac{1}{p_k-1}|G|+\frac{109}{400}|G|\leq\frac{209}{400}|G|<\frac{2}{3}$$
(3) $p_1=2$, $G$ has a normal $2$-complement and $N_2<\frac{|G|}{2}$
Let $H$ be the normal $2$-complement. Then $H$ contains all Sylow subgroups for primes other than $2$. Suppose $H$ is of order $2^{m-a_1}|G|$, where $m<a_1$. By induction on $n$ we have that $$\sum_{k\neq 2}{N_{p_k}}\leq \frac{2}{3}|H|=\frac{2^{m+1}}{3\cdot 2^{a_1}}|G|\leq\frac{1}{3}|G|$$ Since $N_2<\frac{|G|}{2}$ we have that $N_2\leq \frac{|G|}{4}$, hence $$N\leq N_2+\frac{1}{3}|G|\leq \frac{1}{4}|G|+\frac{1}{3}|G|=\frac{7}{12}|G|<\frac{2}{3}|G|$$
(4) $p_1=2$, $a_1\leq 3$, the Sylow $2$-subgroups are abelian but not cyclic, and $G$ does not have a normal $2$-complement
The Walter theorem states that if $O(G)$ is the maximal normal subgroup of odd order, then $G/O(G)$ is a direct product of $2$-groups, projective special linear groups, Janko groups, and Ree groups of type ${}^2G_2$. Let $H$ be a Sylow $2$-subgroup. Then the number of distinct conjugates of $HO(G)$ is the same as the number of Sylow $2$-subgroups of $G/O(G)$. Call this number $N_2(G/O(G))$ and let $N_2(HO(G))$ denote the number of Sylow $2$-subgroups of $HO(G)$. Then $$N_2\leq N_2(G/O(G))N_2(HO(G))$$ We split into cases:
(4a) The Sylow $2$-subgroup of $G/O(G)$ is normal and $[G:HO(G)]>3$
In this case $HO(G)$ contains all Sylow $2$-subgroups. If $HO(G)$ is of index larger than $3$ then $N_2\leq \frac{|G|}{16}$ and we may argue as in case (2).
(4b) The Sylow $2$-subgroup of $G/O(G)$ is normal and $[G:HO(G)]=3$
If $HO(G)$ is of index $3$, then $HO(G)$ contains all Sylow subgroups except for the $3$-subgroups. Let $k$ be the smallest prime such that $p_k$ has an exponent of $1$ in the prime factorization of $|HO(G)|$. We can argue as in (2) to obtain that $$\sum_{p_i\neq 3}{N_{p_i}}=N(HO(G))\leq\frac{23}{50}(|G|/3)+\frac{1}{p_k-1}(|G|/3)<\frac{1}{3}|G|$$ and since $N_{3}\leq \frac{1}{3}|G|$ we have that $N\leq\frac{2}{3}|G|$.
(4c) $G/O(G)$ has a simple projective special linear group as a normal subgroup
In this case the normal subgroup of odd index given by Walter's theorem is either $PSL_2(2^m)$ with $m>1$, $PSL_2(q)$ where $q\equiv \pm 3\pmod{8}$, or either of these groups with an extra $\mathbb{Z}_2$ factor (this follows from the constraint $a_1\leq 3$). By Sylow 2-subgroups of the group $\mathrm{PSL}(2,q)$, $N_2(G/O(G))\leq\frac{1}{12}|G/O(G)|$. If $p_2>3$ or $p_2=3$ and $a_2\geq 2$, then we may argue as in case (2).
If $p_2=3$ and $a_2=1$, we can squeeze out the result by noting that $|O(G)|$ is not divisible by $3$, hence if $H'$ is a Sylow $3$-subgroup then $$N_3\leq N_3(G/O(G))N_3(H'O(G))$$ with the notation having the obvious meaning, hence since the normalizer of the $3$-subgroup has order at least $6$ we have $$\frac{N_3}{|G|}\leq\frac{N_3(G/O(G))}{|G/O(G)|}\leq \frac{1}{6}$$ so $$N\leq\sum_{p>5}{\frac{1}{p^2}}|G|+N_2+N_3+\sum_{p_i\geq 5}{N_{p_i}}\leq\frac{23}{50}|G|-\frac{1}{4}|G|-\frac{1}{9}|G|-\frac{1}{25}|G|+\frac{1}{12}|G|+\frac{1}{4}|G|+\frac{1}{6}|G|=\frac{503}{900}|G|<\frac{2}{3}|G|$$
(4d) $G/O(G)$ has a Janko group $J_1$ as a normal subgroup
In this case the normalizer of a Sylow $2$-subgroup has index $1045$, so $N_2\leq\frac{1}{1045}|G|$, which is more than small enough to apply the argument in case (2).
(4e) $G/O(G)$ has a Ree group as a normal subgroup
I couldn't find a reference on the number of Sylow $2$-subgroups of Ree groups, but note that we can deduce that $N_2\leq \frac{|G|}{24}$ because the Sylow $2$-subgroup is of order $8$ and by Burnside's transfer theorem its normalizer must have an additional factor of at least $3$, and this is enough to apply the argument in case (2).
(5) $p_1=2$, $a_2=3$ and $G$ does not have a normal $2$-complement
By the Frobenius normal $p$-complement theorem we must have that $|N_G(H)|\geq 16$ where $H$ is a Sylow $2$-subgroup, hence $N_2\leq\frac{1}{16}|G|$ and we may argue as in case (2).
(6) $p_1=2$, $a_1=1$ and $N_2=\frac{|G|}{2}$
In this case we are in the situation of Suppose that half of the elements of G have order 2 and the other half form a subgroup H of order n. Prove that H is an abelian subgroup of G., meaning that we have an abelian subgroup of index $2$, hence every Sylow $p_i$ subgroup for $i>1$ is normal, hence $$N=\frac{1}{2}|G|+n-1$$ and since in particular $n-1\leq\frac{|G|}{6}$ because $|G|\geq 2\cdot 3^{n-1}$ we have that $$N\leq \frac{2}{3}|G|$$ This bound is tight, with $S_3$ giving an example of equality (in fact $S_3$ is unique in this sense, which can be seen from the proof).
Best Answer
This is pretty obvious if $H$ contains a Sylow $p$-subgroup of $G$, since then the Sylow $p$-subgroups of $H$ are just the Sylow $p$-subgroups of $G$ contained in $H$, so $n_p(H) \leq n_p(G)$ since $\operatorname{Syl}_p(H) \subseteq \operatorname{Syl}_p(G)$.
If $H$ is smaller though, then something strange seems possible: a Sylow $p$-subgroup of $G$ can have a ton of smaller $p$-subgroups, perhaps $H$ could conspire to have all those smaller $p$-subgroups as Sylow subgroups. For instance, the alternating group of order 12 has only one Sylow 2-subgroup, but it has 3 subgroups of order 2. Why can't a subgroup have those 3 smaller subgroups as its Sylows?
The answer is Lagrange's theorem. If $Q_1, Q_2 \leq P$ are two $p$-subgroups of a Sylow $p$-subgroup $P$ and $Q_1, Q_2 \leq H$ are both contained in $H$, then of course the subgroup $Q=\langle Q_1 ,Q_2 \rangle$ generated by these two $p$-subgroups is contained in both $P$ and $H$. Since $Q \leq P$, it is a $p$-subgroup, and since $Q \leq H$, it is a $p$-subgroup contained in $H$. If $Q_1 \neq Q_2$, then $|Q| > \max(|Q_1|,|Q_2|)$, and so neither $Q_1$ nor $Q_2$ can be a Sylow $p$-subgroup of $H$. In particular, a Sylow $p$-subgroup $P$ of $G$ can only contribute at most one Sylow $p$-subgroup of $H$, namely $H \cap P$.
Not all $H\cap P$ are actually Sylow $p$-subgroups of $H$, but every Sylow $p$-subgroup of $H$ is contained in one of these by Sylow's containment/extension theorem: every $p$-subgroup of $G$ (such as a Sylow $p$-subgroup of $H$) is contained in a Sylow $p$-subgroup of $G$.