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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?

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

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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 $p$.

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    $\begingroup$ 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?). $\endgroup$ Commented Jun 21 at 8:32
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    $\begingroup$ 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/… $\endgroup$ Commented Jun 21 at 9:58
  • $\begingroup$ The book and exercise can be found here bit.ly/4eBePU7, still not clear what is meant. $\endgroup$ Commented Jun 21 at 10:08
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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: \begin{equation} 1\equiv m\equiv \sum_i\sum_j a_{ij}\equiv \sum_j\sum_i a_{ij}\equiv n\pmod{p} \end{equation}

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 $V$.

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    $\begingroup$ Nice @Steve D! +1 from me! This resembles the proofs in the second remark above, see link there. $\endgroup$ Commented Jun 21 at 16:49
  • $\begingroup$ 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. $\endgroup$
    – Damalone
    Commented Jun 22 at 23:27
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    $\begingroup$ $K$ 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. $\endgroup$
    – Steve D
    Commented Jun 23 at 3:17

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