10
$\begingroup$

Let $|G|=p^{n}m$, and let $0\leq k\leq n$. Then the number of subgroups of $G$ of order $p^k$ is congruent to $1$ modulo $p$.

I know Wielandt's proof in the case $k=n$, which is standard, but I dont know how to use that proof to prove this above modification. Clearly the set $S$ on which we should make $G$ act by left multiplication has to be the set of all subsets of size $p^k$ and then we know that $\displaystyle \binom{p^{k}r} {p^{k}}$ is congruent to $r \bmod p$. But somehow using the Orbit-Stabilizer theorem I am unable to prove the above modification.

Could someone help me prove the above statement using a modification of Wielandt's proof?

$\endgroup$
  • $\begingroup$ You might want to check also this [entry][1] [1]: math.stackexchange.com/questions/203850/… $\endgroup$ – Nicky Hekster Aug 30 '13 at 17:42
  • $\begingroup$ I would say, this is Frobenius theorem (1895), rather than Sylow (1872). The reason is, this theorem appears in a paper of Frobenius, whose title is generalization of Sylow theorems and one of the generalization is the theorem you stated. It is not so easy to derive generalization from Sylow's original (third) theorem. see also my answer for similar question math.stackexchange.com/questions/1586060/… $\endgroup$ – p Groups Dec 30 '15 at 12:27
15
$\begingroup$

None of the versions of Wielandt's proofs of Sylow's Theorem that I have seen assume that $k=n$. I will cut and paste the proof from some lecture notes I have.

Theorem. Let $G$ be a finite group and let $p^\beta$ divide $|G|$, where $p$ is prime. Let $k$ be the number of subgroups of $G$ of order $p^\beta $. Then $k \equiv 1 \pmod{p} $.

Proof. Let $|G| = p^\alpha t$ with $p\nmid t$, so $\beta \le \alpha$.

Let $\Omega $ be the set of all subsets of $G$ of order $p^\beta $. So $|\Omega | = \left( \begin{array}{c} p^\alpha t \\ p^\beta \end{array} \right)$.

Let $G$ act on $\Omega $ by right multiplication; i.e. if $S \in \Omega$, then $S^g = Sg = \{sg \mid s \in S\}$.

Let $\Gamma$ be an orbit of $G^\Omega$. If $T \in \Gamma$ and $x\in T$ then $1\in Tx^{-1} \in \Gamma$ so there is a set $S\in \Gamma$ with $1\in S$.

Consider the stabilizer, ${\rm Stab} _G(S)$. If $g\in {\rm Stab} _G(S)$ then $Sg=S$ so $1g = g \in S$. Thus ${\rm Stab} _G(S) \subseteq S$.

1) Suppose that ${\rm Stab} _G(S) = S$, so $S$ is a subgroup of $G$. Then, by the Orbit-Stabilizer Theorem, $|\Gamma| = |G|/|{\rm Stab} _G(S)| = p^{\alpha }t/p^{\beta } = p^{\alpha -\beta }t $ and $\Gamma$ is the set of right cosets of $S$ in $G$. Thus only one element of $\Gamma $ is a subgroup. Conversely, if $T$ is a subgroup of $G$ of order $p^{\beta}$ then $T^G$ (the orbit of $G^\Omega$ containing $T$) is the set of right cosets of $T$ in $G$ so has length $p^{\alpha -\beta }t $

2) Suppose that ${\rm Stab} _G(S) \ne S$. Then $|S| > |{\rm Stab} _G(S)|$, so $|\Gamma | > p^{\alpha -\beta }t $. Since $|\Gamma |$ divides $|G| = p^{\alpha }t $ we have $p^{\alpha -\beta +1}$ divides $|\Gamma |$. So by 1) no element of $\Gamma$ is a subgroup of $G$ in this case.

Hence, there are exactly $k$ orbits whose stabilizer has size $p^\beta $ and these orbits have have length $p^{\alpha -\beta }t $, whereas those orbits whose stabilizer has size less than $p^\beta $ have length divisible by $p^{\alpha -\beta +1} $.

So $|\Omega | = kp^{\alpha -\beta }t + l p^{\alpha -\beta +1} $ for some $l$, and hence $$|\Omega |/p^{\alpha -\beta } = kt + lp \equiv kt \pmod{p}.$$ Since $p \nmid t$, there is a unique $u \in \{1,\ldots,p-1\}$ with $ut \equiv 1 \pmod{p}$, and multiplying by $u$ gives $$k \equiv |\Omega |u/p^{\alpha -\beta } \!\!\!\pmod{p} \equiv \left( \begin{array}{c} p^\alpha t \\ p^\beta \end{array} \right)u /p^{\alpha -\beta }\!\!\! \pmod{p}.$$

It is possible to prove directly that this last expression equals 1 mod $p$, but we can avoid that as follows. Note that $k \pmod{p}$ is a function of $|G|$ and $p^\beta$ only, and so it is the same for all groups of order $p^{\alpha} t$. So $k \pmod{p}$ can be determined from $G=C_{p^{\alpha} t}$, the cyclic group of order $p^{\alpha} t$.

Hence $k \equiv 1 \pmod{p}$, since a cyclic group $G$ has a unique subgroup of each order dividing $|G|$.

$\endgroup$
  • $\begingroup$ Wait, if I understood correctly, your $k$ is the number of orbits that contains a subgroup (of order $p^\beta$), but should that not depend on the group structure? I mean, the way that the stabilizer behaves should depend on $G$? $\endgroup$ – user134070 Mar 27 '16 at 10:27
  • $\begingroup$ For instant, this can be applied to $\beta=k$, in which case the groups are Sylow $p$-subgroups, but abelian groups always have a unique Sylow $p$-subgroups, whereas non-abelian groups not necessarily. $\endgroup$ – user134070 Mar 27 '16 at 10:28
  • 1
    $\begingroup$ Yes of course $k$ depends on $G$. It is the value of $k$ modulo $p$ that depends only on $|G|$ and not on the structure of $G$. $\endgroup$ – Derek Holt Mar 27 '16 at 11:41
  • $\begingroup$ Very nice solution, especially last argument using cyclic groups! $\endgroup$ – Mikhail Goltvanitsa Dec 13 '17 at 20:27

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.