# Generalized Sylow's theorem [closed]

I'm working through some exercises in Alperin and Bell's textbook "Groups and Representations." I came across a very interesting exercise which generalizes Sylow's theorem:

Ex 7.4: If $$|G|$$ is divisible by $$p^b$$, and $$H \leq G$$ has order $$p^a$$ where $$a \leq b$$, then the number of subgroups of $$G$$ that both contain $$H$$ and have order $$p^b$$ is congruent to 1 modulo $$p$$.

This is quite a generalization of Sylow's theorems; even taking $$H$$ to be the trivial group is already quite interesting. But I don't know how to even get started on this. Does anyone know how to solve this problem?

• This is a result of [E. Spiegel, Another look at Sylow’s third theorem, Math. Mag. 77 (2004), 227–232] Commented Jun 21 at 17:10

Let me sketch you a proof. I assume you are familiar with the Sylow Theorems. Let us fix $$Q$$, a $$p$$-subgroup of $$G$$, and let $$Q$$ act by conjugation on the set $$\Omega$$ of all Sylow $$p$$-subgroups of $$G$$. Observe that $$\#\Omega \equiv 1$$ mod $$p$$. We are going to apply the Orbit-Stabilizer Theorem, so we need to inspect the orbit lengths. To this end we need a lemma.

Lemma If $$P \in Syl_p(G)$$, then $$Q \cap N_G(P)=P \cap Q$$

Proof $$N_G(P) \cap Q$$ is a $$p$$-subgroup of $$N_G(P)$$, it must be contained in some Sylow $$p$$-subgroup of $$N_G(P)$$, but $$N_G(P)$$ contains only one Sylow $$p$$-subgroup, namely $$P$$ (being normal in $$N_G(P)$$). Hence $$N_G(P) \cap Q \subseteq P$$, so $$N_G(P) \cap Q \subseteq P \cap Q$$ and the reverse inclusion is trivial. $$\square$$

It follows from the lemma that the length of the orbit of a Sylow $$p$$-subgroup under the action of $$Q$$ on $$\Omega$$ is $$|Q:N_Q(P)|=|Q:P \cap Q|$$. Hence the orbits of length $$1$$ correspond exactly with those $$P$$ such that $$Q \subseteq P$$. The rest of the orbits have lengths divisible by $$p$$. And this gives you the required result by taking the Orbit-Stabilizer formula mod $$p$$.

• Doesn't this only prove the desired result if $b$ is the largest power of $p$ that divides $|G|$? I was interpreting $b$ to be arbitrary but maybe that's not what was intended (maybe it's not true?). Commented Jun 21 at 8:32
• Hi @QiaochuYuan, nice to see you again on MSE! Yes you are right, I was interpreting the power $p^b$ as the largest power of $p$ dividing $|G|$. If that is not the case, then it is much more complicated, see math.stackexchange.com/questions/2952262/… Commented Jun 21 at 9:58
• The book and exercise can be found here bit.ly/4eBePU7, still not clear what is meant. Commented Jun 21 at 10:08

Here is a proof based on a slightly easier result (where $$H=1$$) that I learned from a book by Dickson IIRC.

We can prove this inductively, the case $$b=a$$ being trivial. For the inductive step, build an $$m\times n$$ matrix, where the rows are indexed by the $$m$$ subgroups of order $$p^b$$ containing $$H$$, and the columns are indexed by the $$n$$ subgroups of order $$p^{b+1}$$ containing $$H$$. Call this matrix $$A$$, and let $$a_{ij}$$ be equal to $$1$$ if the corresponding column group contains the corresponding row group, and $$a_{ij}=0$$ otherwise. By induction, we're assuming $$m\equiv1\pmod{p}$$.

For every row group $$K$$, there are $$1\pmod{p}$$ column groups containing it. This is because $$K$$ is normal in each of them, and so we're just counting subgroups of order $$p$$ in $$N_G(K)/K$$. For every column group, there are $$1\pmod{p}$$ row groups that it contains. This is equivalent to counting the maximal subgroups that contain $$H$$, and a proof that there are $$1\pmod{p}$$ of them is given below.

Now we just add up the entries of the matrix in two different ways: $$$$1\equiv m\equiv \sum_i\sum_j a_{ij}\equiv \sum_j\sum_i a_{ij}\equiv n\pmod{p}$$$$

We still need to prove that for a group $$P$$ of order $$p^{b+1}$$ containing $$H$$, it has $$1\pmod{p}$$ maximal subgroups containing $$H$$. Since $$\Phi(P)$$ is contained in every maximal subgroup, this is equivalent to counting the maximal subgroups that contain $$H\Phi(P)$$. The key difference is this latter subgroup is normal in $$P$$, so we can count maximal subgroups in $$P/H\Phi(P)$$, which is an $$\mathbb{F}_p$$ vector space. It is well-known (and I can provide proof if need be) that the number of codimension-1 subpsaces of a finite $$\mathbb{F}_p$$ vector space is $$1\pmod{p}$$.

EDIT: Here is a proof of the codimension-1 claim above.

Assume $$V$$ has dimension $$d$$ over the finite field $$\mathbb{F}_p$$. Define the set $$\begin{equation*} X_V=\{(v_1,\ldots,v_{d-1})\in V^{d-1}\mid\dim(\langle v_1,\ldots,v_{d-1}\rangle)=d-1\} \end{equation*}$$ of linearly independent lists from $$V$$ of length $$d-1$$. If $$\mathcal{M}=\{M\le V\mid\dim(M)=d-1\}$$ is the set of maximal subgroups (subspaces), there is a surjective map $$f:X_V\rightarrow\mathcal{M}$$ given by $$\begin{equation*} (v_1,\ldots,v_{d-1})\mapsto\langle v_1,\ldots,v_{d-1}\rangle \end{equation*}$$ For a fixed $$M\in\mathcal{M}$$, note that $$f^{-1}(M)=X_M$$, the set of bases of $$M$$. Since $$|X_M|$$ only depends on $$\dim(M)$$, we see that $$|f^{-1}(M)|=|f^{-1}(N)|$$ for all $$M,N\in\mathcal{M}$$, and hence $$\begin{equation*} |\mathcal{M}| = \frac{|X_V|}{|X_M|} = \frac{\prod\limits_{i=0}^{d-2}(p^d-p^i)}{\prod\limits_{i=0}^{d-2}(p^{d-1}-p^i)} = \frac{p^d-1}{p-1}\equiv1\pmod{p} \end{equation*}$$

The above proof works similarly (with $$k$$ replacing $$d-1$$) to count the $$k$$-dimensional subspaces of $$V$$.

• Nice @Steve D! +1 from me! This resembles the proofs in the second remark above, see link there. Commented Jun 21 at 16:49
• Thanks! Could you explain why the row group $K$ is normal in the column groups containing it? Also, what is $\Phi(P)$? Finally, it would be great if you could provide the proof of the number of codimension-1 subspaces. Commented Jun 22 at 23:27
• $K$ is normal in each of the column groups because it has index $p$ in them, and normalizers grow. $\Phi(P)$ is the Frattini subgroup of $P$, the intersection of all maximal subgroups. I'll edit in a proof of the codimension-1 stuff. Commented Jun 23 at 3:17