*agrégation*lesson on finite fields this afternoon, and I could not resist having the student check this. Then came the natural question of describing the Sylow subgroups of this finite group. This is what I describe here.

First of all, let's recall the computation of the cardinality of $G$. The first column of a matrix in $G$ must be non-zero, hence there are $7^2-1$ possibilities; for the second column, it only needs to be non-collinear to the first one, and each choice of the first column forbids $7$ second columns, hence $7^2-7$ possibilities. In the end, one has $\mathop{\rm Card}(G)=(7^2-1)(7^2-7)=48\cdot 42=2016$. The same argument shows that the cardinality of the group $\mathop{\rm GL}(n,\mathbf F_q)$ is equal to $(q^n-1)(q^n-q)\cdots (q^n-q^{n-1})=q^{n(n-1)/2}(q-1)(q^2-1)\cdots (q^n-1)$.

Let's go back to our example. The factorization of this cardinal comes easily: $2016=(7^2-1)(7^2-7)=(7-1)(7+1)7(7-1)=6\cdot 8\cdot 7\cdot 6= 2^5\cdot 3^2\cdot 7$. Consequently, there are three Sylow subgroups to find, for the prime numbers $2$, $3$ and $7$.

The cas $p=7$ is the most classical one. One needs to find a group of order 7, and one such subgroup is given by the group of upper triangular matrices $\begin{pmatrix} 1 & * \\ 0 & 1\end{pmatrix}$. What makes things work is that $p$ is the characteristic of the chosen finite field. In general, if $q$ is a power of $p$, then the subgroup of upper-triangular matrices in $\mathop{\rm GL}(n,\mathbf F_q)$ with $1$s one the diagonal has cardinality $q\cdot q^2\cdots q^{n-1}=q^{n(n-1)/2}$, which is exactly the highest power of $p$ divising the cardinality of $\mathop{\rm GL}(n,\mathbf F_q)$.

Let's now study $p=3$. We need to find a group $S$ of order $3^2=9$ inside $G$. There are a priori two possibilities, either $S\simeq (\mathbf Z/3\mathbf Z)^2$, or $S\simeq (\mathbf Z/9\mathbf Z)$.

We will find a group of the first sort, which will that the second case doesn't happen, because all $3$-Sylows are pairwise conjugate, hence isomorphic.

Now, the multiplicative group $\mathbf F_7^\times$ is of order $6$, and is cyclic, hence contains a subgroup of order $3$, namely $C=\{1,2,4\}$. Consequently, the group of diagonal matrices with coefficients in $C$ is isomorphic to $(\mathbf Z/3\mathbf Z)^2$ and is our desired $3$-Sylow.

Another reason why $G$ does not contain a subgroup $S$ isomorphic to $\mathbf Z/9\mathbf Z$ is that it does not contain elements of order $9$. Let's argue by contradiction and consider a matrix $A\in G$ such that $A^9=I$; then its minimal polynomial $P$ divides $T^9-1$. Since $7\nmid 9$, the matrix $A$ is diagonalizable over the algebraic closure of $\mathbf F_7$. The eigenvalues of $A$ are eigenvalues are $9$th roots of unity, and are quadratic over $\mathbf F_7$ since $\deg(P)\leq 2$. On the other hand, if $\alpha$ is a $9$th root of unity belonging to $\mathbf F_{49}$, one has $\alpha^9=\alpha^{48}=1$, hence $\alpha^3=1$ since $\gcd(9,48)=3$. Consequently, $\alpha$ is a cubic root of unity and $A^3=1$, showing that $A$ has order $3$.

It remains to treat the case $p=2$, which I find slightly trickier. Let's try to find elements $A$ in $G$ whose order divides $2^5$. As above, it is diagonalizable in an algebraic closure, its minimal polynomial divides $T^{32}-1$, and its roots belong to $\mathbf F_{49}$, hence satisfy $\alpha^{32}=\alpha^{48}=1$, hence $\alpha^{16}=1$. Conversely, $\mathbf F_{49}^\times$ is cyclic of order $48$, hence contains an element of order $16$, and such an element is quadratic over $\mathbf F_7$, hence its minimal polynomial $P$ has degree $2$. The corresponding companion matrix $A$ in $G$ is an element of order $16$, generating a subgroup $S_1$ of $G$ isomorphic to $\mathbf Z/16\mathbf Z$. We also observe that $\alpha^8=-1$ (because its square is $1$); since $A^8$ is diagonalizable in an algebraic closure with $-1$ as the only eigenvalue, this shows $A^8=-I$.

Now, there exists a $2$-Sylow subgroup containing $S_1$, and $S_1$ will be a normal subgroup of $S$ (because its index is the smallest prime number dividing the order of $S$, which is $2$). This suggests to introduce the normalizer $N$ of $S_1$ in $G$. One then has $S_1\subset S\subset N$. Let $s\in S$ be such that $s\not\in S_1$; then there exists a unique $k\in\{1,\dots,15\}$ such that $s^{-1}As=A^k$, and $s^{-2}As^2=A^{k^2}=A$ (because $s$ has order $2$ modulo $S_1$), hence $k^2\equiv 1\pmod{16}$—in other words, $k\equiv \pm1\pmod 8$.

There exists a natural choice of $s$: the involution ($s^2=I$) which exchanges the two eigenspaces of $A$. To finish the computation, it's useful to take a specific example of polynomial $P$ of degree $2$ whose roots in $\mathbf F_{49}$ are primitive $16$th roots of unity. In other words, we need to factor the $16$th cyclotomic polynomial $\Phi_{16}=T^8+1$ over $\mathbf F_7$ and find a factor of degree $2$; actually, Galois theory shows that all factors have the same degree, so that there should be 4 factors of degree $2$. To explain the following computation, some remark is useful. Let $\alpha$ be a $16$th root of unity in $\mathbf F_{49}$; we have $(\alpha^8)^2=1$ but $\alpha^8\neq 1$, hence $\alpha^8=-1$. If $P$ is the minimal polynomial of $\alpha$, the other root is $\alpha^7$, hence the constant term of $P$ is equal to $\alpha\cdot \alpha^7=\alpha^8=-1$.

We start from $T^8+1=(T^4+1)^2-2T^4$ and observe that $2\equiv 4^2\pmod 7$, so that $T^8+1=(T^4+1)^2-4^2T^4=(T^4+4T^2+1)(T^4-4T^2+1)$. To find the factors of degree $2$, we remember that their constant terms should be equal to $-1$. We thus go on differently, writing $T^4+4T^2+1=(T^2+aT-1)(T^2-aT-1)$ and solving for $a$: this gives $-2-a^2=4$, hence $a^2=-6=1$ and $a=\pm1$. The other factors are found similarly and we get

\[ T^8+1=(T^2-T-1)(T^2+T-1)(T^2-4T-1)(T^2+4T-1). \]

We thus choose the factor $T^2-T-1$ and set $A=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$.

Two eigenvectors for $A$ are $v=\begin{pmatrix} 1 \\ \alpha \end{pmatrix}$ and $v'=\begin{pmatrix}1 \\ \alpha'\end{pmatrix}$, where $\alpha'=\alpha^7$ is the other root of $T^2-T-1$. The equations for $B$ are $Bv=v'$ and $Bv'=v$; this gives $B=\begin{pmatrix} 1 & 0 \\ 1 & - 1\end{pmatrix}$. The subgroup $S=\langle A,B\rangle$ generated by $A$ and $B$ has order $32$ and is a $2$-Sylow subgroup of $G$.

Generalizing this method involves finding large commutative $p$-subgroups (such as $S_1$) which belong to appropriate (possibly non-split) tori of $\mathop{\rm GL}(n)$ and combining them with adequate parts of their normalizer, which is close to considering Sylow subgroups of the symmetric group. The paper Sylow $p$-subgroups of the classical groups over finite fields with characteristic prime to $p$ by A.J. Weir gives the general description (as well as for orthogonal and symplectic groups), building on an earlier paper in which he constructed Sylow subgroups of symmetric groups. See also the paper Some remarks on Sylow subgroups of the general linear groups by C. R. Leedham-Green and W. Plesken which says a lot about maximal $p$-subgroups of the general linear group (over non-necessarily finite fields). Also, the question was recently the subject of interesting discussions on MathOverflow.

[Edited on Febr. 14 to correct the computation of the 2-Sylow...]

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