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## Meditation on the Sylow theorems II

In Part I we discussed some conceptual proofs of the Sylow theorems. Two of those proofs involve reducing the existence of Sylow subgroups to the existence of Sylow subgroups of $S_n$ and $GL_n(\mathbb{F}_p)$ respectively. The goal of this post is to understand the Sylow $p$-subgroups of $GL_n(\mathbb{F}_p)$ in more detail and see what we can learn from them about Sylow subgroups in general.

Explicit Sylow theory for $GL_n(\mathbb{F}_p)$

Our starting point is the following.

Baby Lie-Kolchin: Let $P$ be a finite $p$-group acting linearly on a finite-dimensional vector space $V$ over $\mathbb{F}_p$. Then $P$ fixes a nonzero vector; equivalently, $V$ has a trivial subrepresentation.

Proof. If $\dim V = n$ then there are $p^n - 1$ nonzero vectors in $V \setminus \{ 0 \}$, so by the PGFPT $P$ fixes at least one of them (in fact at least $p-1$ but these are just given by scalar multiplication). $\Box$

Now we can argue as follows. If $P$ is a finite $p$-group acting on an $n$-dimensional vector space $V$ over $\mathbb{F}_p$ (equivalently, up to isomorphism, a finite $p$-subgroup of $GL_n(\mathbb{F}_p)$), it fixes some nonzero vector $v_1 \in V$. Writing $V_1 = \text{span}(v_1)$, we get a quotient representation on $V/V_1$, on which $P$ fixes some nonzero vector, which we lift to a vector $v_2 \in V$, necessarily linearly independent from $v_1$, such that $P$ acts upper triangularly on $V_2 = \text{span}(v_1, v_2)$.

Continuing in this way we get a sequence $v_1, \dots v_n$ of linearly independent vectors (hence a basis of $V$) and an increasing sequence $V_k = \text{span}(v_1, \dots v_k)$ of subspaces of $V$ that $P$ leaves invariant, satisfying the additional condition that $P$ fixes $v_k \bmod V_{k-1}$. The subspaces $V_k$ form a complete flag in $V$, and writing the elements of $P$ as matrices with respect to the basis $v_1, \dots v_n$, we see that the conditions that $P$ leaves $V_k$ invariant and fixes $v_k \bmod V_{k-1}$ says exactly that $P$ acts by upper triangular matrices with $1$s on the diagonal in this basis.

Conjugating back to the standard basis, we’ve proven:

Proposition: Every $p$-subgroup of $GL_n(\mathbb{F}_p)$ is conjugate to a subgroup of the unipotent subgroup $U_n(\mathbb{F}_p)$.

This is almost a proof of Sylow I and II for $GL_n(\mathbb{F}_p)$ (albeit at the prime $p$ only), but because we defined Sylow $p$-subgroups to be $p$-subgroups having index coprime to $p$, we’ve only established that $U_n(\mathbb{F}_p)$ is maximal, not that it’s Sylow.

We can show that it’s Sylow by explicitly dividing its order into the order of $GL_n(\mathbb{F}_p)$ but there’s a more conceptual approach that will teach us more. Previously we proved the normalizer criterion: a $p$-subgroup $P$ of a group $G$ is Sylow iff the quotient $N_G(P)/P$ has no elements of order $p$.

Claim: The normalizer of $U = U_n(\mathbb{F}_p)$ is the Borel subgroup $B = B_n(\mathbb{F}_p)$ of upper triangular matrices (with no restrictions on the diagonal). The quotient $B/U$ is the torus $(\mathbb{F}_p^{\times})^n$ and in particular has no elements of order $p$.

Corollary (Explicit Sylow I and II for $GL_n(\mathbb{F}_p)$: $U_n(\mathbb{F}_p)$ is a Sylow $p$-subgroup of $GL_n(\mathbb{F}_p)$.

Proof. The normalizer $N_G(U)$ is the stabilizer of $G = GL_n(\mathbb{F}_p)$ acting on the set of conjugates of $U$. We want to show that $N_G(U) = B$, which would mean that the action of $G$ on the conjugates of $U$ can be identified with the quotient $G/B$.

This quotient is the complete flag variety: it can be identified with the action of $G$ on the set of complete flags in $\mathbb{F}_p^n$, since the action on flags is transitive and the stabilizer of the standard flag

$\displaystyle 0 \subset \text{span}(e_1) \subset \text{span}(e_1, e_2) \subset \dots \subset \mathbb{F}_p^n$

is exactly $B$. So it suffices to exhibit a $G$-equivariant bijection between conjugates of $U$ and complete flags which sends $U$ to the standard flag, since then their stabilizers must agree.

But this is clear: given a complete flag

$\displaystyle 0 = V_0 \subset V_1 \subset \dots \subset V_n = \mathbb{F}_p^n$

we can consider the subgroup of $g \in G$ which preserves the flag (so $g V_i = V_i$) and which has the additional property that the induced action on $V_{i+1}/V_i$ is trivial for every $i$. This produces $U$ when applied to the standard flag, so produces conjugates of $U$ when applied to all flags. In the other direction, given a conjugate $gUg^{-1}$ of $U$, it has a $1$-dimensional invariant subspace $V_1$ acting on $\mathbb{F}_p^n$, quotienting by this subspace produces a unique $1$-dimensional invariant subspace $V_2/V_1$ acting on $\mathbb{F}_p^n/V_1$, etc.; this produces the standard flag when applied to $U$, so produces $g$ applied to the standard flag when applied to conjugates of $U$. So we get our desired $G$-equivariant bijection between conjugates of $U$ and complete flags, establishing $N_G(U) = B$ as desired. $\Box$

(This argument works over any field.)

From here it’s not hard to also prove

Explicit Sylow III for $GL_n(\mathbb{F}_p)$: The number $n_p$ of conjugates of $U_n(\mathbb{F}_p)$ in $GL_n(\mathbb{F}_p)$ divides the order of $GL_n(\mathbb{F}_p)$ and is congruent to $1 \bmod p$.

Proof. Actually we can compute $n_p$ exactly: we established above that it’s the number of complete flags in $\mathbb{F}_p^n$ (on which $GL_n(\mathbb{F}_p)$ acts transitively, hence the divisibility relation), and a classic counting argument (count the number of possibilities for $V_1$, then for $V_2$, etc.) gives the $p$-factorial

$\displaystyle n_p = [n]_p! = \prod_{i=1}^n [i]_p = \prod_{i=1}^n \left( p^{i-1} + p^{i-2} + \dots + 1 \right)$

which is clearly congruent to $1 \bmod p$. $\Box$

We could also have done this by dividing the order of $GL_n(\mathbb{F}_p)$ by the order of the Borel subgroup $B$, but again, doing it this way we learn more, and in fact we get an independent proof of the formula

$\displaystyle |GL_n(\mathbb{F}_p)| = p^{ {n \choose 2} } (p - 1)^n [n]_p!$

for the order of $GL_n(\mathbb{F}_p)$, where all three factors acquire clear interpretations: the first factor is the order of the unipotent subgroup $U$, the second factor is the order of the torus $B/U \cong (\mathbb{F}_p^{\times})^n$, and the third factor is the size of the flag variety $G/B$.

What is going on in these proofs?

Let’s take a step back and compare these explicit proofs of the Sylow theorems for $GL_n(\mathbb{F}_p)$ to the general proofs of the Sylow theorems. The first three proofs we gave of Sylow I (reduction to $S_n$, reduction to $GL_n(\mathbb{F}_p)$, action on subsets) all proceed by finding some clever way to get a finite group $G$ to act on a finite set $X$ with the following two properties:

• $|X| \not \equiv 0 \bmod p$. By the PGFPT this means any $p$-subgroups of $G$ act on $X$ with fixed points, so we can look for $p$-subgroups in the stabilizers $\text{Stab}_G(x), x \in X$.
• The stabilizers of the action of $G$ on $X$ are $p$-groups. Combined with the first property, this means that at least one stabilizer must be a Sylow $p$-subgroup, since at least one stabilizer must have index coprime to $p$.

In fact finding a transitive such action is exactly equivalent to finding a Sylow $p$-subgroup $P$, since it must then be isomorphic to the action of $G$ on $G/P$. The nice thing, which we make good use of in the proofs, is that we don’t need to restrict our attention to transitive actions, because the condition that $|X| \not \equiv 0 \bmod p$ has the pleasant property that if it holds for $X$ then it must hold for at least one of the orbits of the action of $G$ on $X$. (This is a kind of “$\bmod p$ pigeonhole principle.”)

In the explicit proof for $G = GL_n(\mathbb{F}_p)$ we find $X$ by starting with the action of $G$ on the set of nonzero vectors $\mathbb{F}_p^n \setminus \{ 0 \}$, which satisfies the first condition but not the second, and repeatedly “extending” the action (to pairs of a nonzero vector $v_1$ and a nonzero vector in the quotient $\mathbb{F}_p^n/v_1$, etc.) until we arrived at an action satisfying both conditions, namely the action of $G$ on the set of tuples of a nonzero vector $v_1$, a nonzero vector $v_2 \in \mathbb{F}_p^n/v_1$, a nonzero vector $v_3 \in \mathbb{F}_p^n / \text{span}(v_1, v_2)$, etc. (a slightly decorated version of a complete flag).

The next question we’ll address in Part III is: can we do something similar for $G = S_n$?

### One Response

1. Hi t0rajir0u!