The Klein group, the centralizer of a permutation, and its relation with the alternating group

The following reflexion came out of my irrepressible need to understand why the 3 double transpositions in $\mathfrak S_4$, together with the identity, formed a group $V$. Of course, one might just say: “they are stable under multiplication, as one sees by computing the 4·3/2 = 6 different products”, but who makes this computation anyway? And since I wanted not only to understand this, but to explain it to Lean, I needed an argument that could actually be done, for real. So here is an argument that requires no computation, besides the one that says that there are 3 double transpositions.

Prop. — The subgroup of $\mathfrak S_4$ generated by the 3 double transpositions is the unique 2-sylow subgroup of $\mathfrak A_4$. In particular, it has order 4 and consists of these 3 double transpositions and of the identity.
Proof. — Let $V$ be the subset of $\mathfrak S_4$ consisting of these 3 double transpositions and of the identity. Let $S$ be a 2-sylow subgroup in $\mathfrak A_4$.
We first prove $S \subseteq V$. The subgroup $S$ has order 4. Let $g\in S$. The order of $g$ divides 4, so its cycles have lengths 1, 2 or 4. If there were one cycle of length 4, then $g$ would be that cycle, hence of odd sign. Consequently, either $g=1$, or $g$ has a cycle of length 2, and then there must be a second because $g$ is even. Consequently, $S\subseteq V$, as claimed.
Since $4=\operatorname{\rm Card}(S)=\operatorname{\rm Card}(V)$, this shows that $S=V$, hence $S=\langle V\rangle$.

At this point, we still need to understand why there are 3 double transpositions. More generally, I wanted to prove that the number of permutations in $\mathfrak S_n$ of given orbit type. The orbit type a permutation $g$ is a multiset of strictly positive integers with sum $n$ given by the cardinalities of the orbits of $g$ on $\{1,\dots,n\}$. We write it as $1^{n_1} 2^{n_2}\dots r^{n_r}$, meaning that $g$ has $n_1$ orbits of length $1$ (fixed points), $n_2$ orbits of cardinality $2$, etc., so that $n= \sum n_i i$. Let $\mathscr O_g$ be the set of orbits of $g$. The action of $g$ on a given orbit $c$ coincides with a circular permutation with order the length $\ell(c)$ of this orbit; when it is nontrivial, such a permutation will be called a cycle of $g$. The supports of these cycles are pairwise disjoint, so that these cycles commute, and their product is exactly $g$. In fact, this is the only way of writing $g$ as a product of cycles with pairwise disjoint supports. (By convention, the identity is not a cycle.)

Theorem. — There are exactly \[ N(1^{n_1}\dots r^{n_r}) = \frac{n!}{1^{n_1}\dots r^{n_r} n_1!\dots n_r!} \] permutations with orbit type $1^{n_1} 2^{n_2}\dots r^{n_r}$.

A standard proof of this result goes as follows. Write the decomposition of such a permutation $g$ into cycles with disjoint supports as $g=(\cdot)\dots (\cdot)(\cdot,\cdot)\dots(\cdot,\cdot,\dots)$, leaving blank spaces for the values of the cycles (and, contradicting our convention, allowing for cycles of length 1…). There are $n!$ ways to fill these spaces with the $n$ distinct integers between $1$ and $n$, but some of them will give rise to the same permutation. Indeed, the entries in a cycle of length $s$ only count up to a circular permutation, so that we need to divide the result by $1^{n_1}\dots r^{n_r}$. Moreoveer, we can switch the order of the cycles of given length, hence we also need to divide that result by $n_s!$ (number of ways of switching the various cycles of length $s$), for all possible length $s$.

This is reasonably convincing but one could wish for something more precise, both in the proof, and in the statement. In fact, in the preceding formula, the numerator $n!$ is the order of $\mathfrak S_n$. Since all permutations with a given orbit type are conjugate by $\mathfrak S_n$, the left hand side appears as the cardinality of the orbit of a permutation $g$ of that orbit type, so that the denominator has to be equal the cardinality of the stabilizer of this permutation under the action by conjugation. Therefore, a more precise proof of this formula could run by elucidating the structure of this centralizer. This may also be interesting once one wishes to relativize the result to the alternating group $\mathfrak A_n$ in order to obtain a formula for the cardinality of the various conjugacy classes in $\mathfrak A_n$.

Let us fix a permutation $g\in\mathfrak S_n$ with orbit type $1^{n_1}\dots r^{n_r}$. The stabilizer of $g$ under the action by conjugation is its centralizer $Z_g$, the subgroup of all $k\in\mathfrak S_n$ which commute with $g$.

We first define a morphism of groups \[ \tau \colon Z_g \to \mathfrak S_{n_1}\times \mathfrak S_{n_2}\times\dots \mathfrak S_{n_r}. \] Let $\mathscr O_g$ be the set of orbits of $g$; this is a set with cardinality $n_1+n_2+\dots+n_r$. Restricted to one orbit, the action of $g$ coincides with that of a circular permutation on (which fixes the complementary subset); these circular permuations have disjoint supports, hence they commute pairwise and their product is equal to $g$. For $c\in\mathscr O_g$, we write $\ell(c)$ for its cardinality of its support, this is also the order of the cycle of $g$ defined by this orbit. If $k\in Z_g$, then $kgk^{-1}=g$. Consequently, the action of $k$ permutes the orbits of $g$, respecting their cardinalities. This defines the desired group morphism $\tau$ from $Z_g$ to a product of permutation groups $\mathfrak S_{n_1}\times \dots \mathfrak S_{n_r}$.

This morphism $\tau$ is surjective.
Indeed, given permutations $\sigma_1$ of the set of fixed points of $g$, $\sigma_2$ of the set of orbits of length 2, etc., we construct $k_\sigma\in Z_g$ such that $\tau(k_\tau)=(\sigma_1,\dots,\sigma_r)$. We fix a point $a_c$ in each orbit $c$ and decide that $k_\sigma(a_c)=a_{\sigma_i(c)}$ if $c$ has length $i$. The formula $k_\sigma g=g_\sigma$ imposes $k_\sigma (g^n a_c)=g^n a_{\sigma_i(c)}$ for all $n\in\mathbf Z$, and it remains to check that this formula gives a well defined element in $Z_g$. In fact, this formula defines a group theoretic section of $\tau$.

What is the kernel of this morphism $\tau$?
If $\tau(k)=1$, then $k$ fixes every orbit $c\in\mathscr O_g$. Since $kg=gk$, we obtain that on each orbit $c$, $k$ coincides with some power of the corresponding cycle, which has order $\ell(c)$. We thus obtain an isomorphism \[ \ker(\tau) \simeq \prod_{c\in\mathscr C_g} (\mathbf Z/\ell(c)\mathbf Z). \]

To compute the cardinality of $Z_g$, it is now sufficient to compute those of $\operatorname{\rm im}(\tau)$ and $\ker(\tau)$, and this gives the formula \[ \operatorname{\rm Card} (Z_g) = \operatorname{\rm Card} (\ker(\tau)) \operatorname{\rm Card} (\operatorname{\rm im}(\tau)) = 1^{n_1}\dots r^{n_r} n_1! \dots n_r!, \] as was to be shown.

One of the interest of this argument is that it can be pushed forward to understand the structure of the conjugacy classes in the alternating group $\mathfrak A_n$. The case $n\leq 1$ is uninteresting, hence we assume $n\geq 2$. Then $\mathfrak A_n$ has index 2 in $\mathfrak S_n$, and the formulas \[ \operatorname  {\rm Card}((g)_{\mathfrak A_n}) = \frac{{\rm Card}({\mathfrak A_n})}{{\rm Card}(Z_g \cap {\mathfrak A_n})} \quad\text{\rm and}\quad \operatorname  {\rm Card}((g)_{\mathfrak S_n}) = \frac{{\rm Card}({\mathfrak S_n})}{{\rm Card}(Z_g)} \] for the cardinalities of the conjugacy classes $(g)_{\mathfrak A_n}$ and $(g)_{\mathfrak S_n}$ imply that both are equal if and only if $Z_g$ is not contained in $\mathfrak A_n$; otherwise, the conjugacy class $(g)_{\mathfrak S_n}$ is the disjoint union of $(g)_{\mathfrak A_n}$ and of a conjugacy class $(g')_{\mathfrak A_n}$ of a permutation $g'$ which is conjugate to $g$ in $\mathfrak S_n$ but not in $\mathfrak A_n$, and both have the same cardinality.

Examples of this phenomenon are classical. For example, the 5-cycles in $\mathfrak S_5$ are conjugate, but they constitute two distinct conjugacy classes under $\mathfrak A_5$. Even more elementary, the 3-cycles $(1\,2\,3)$ and $(1\,3\,2)$ are conjugate in $\mathfrak S_3$, but they are not conjugate in $\mathfrak A_3$ since that group is commutative!

So let us use our description of $Z_g$ to give a full description of this phenomenon.

As a first step, when is $\ker(\tau)$ contained in $\mathfrak A_n$? We have seen that $\ker(\tau)$ is generated by the cycles $c\in\mathscr C_g$. Consequently, $\ker(\tau)$ is contained in $\mathfrak A_n$ if and only if all of them are contained in $\mathfrak A_n$, which means that their lengths are odd.

We assume that this condition holds, so that $\ker(\tau)\subseteq \mathfrak A_n$, and now work on the image of $\tau$. Its surjectivity was proved by the means of an explicit section $\sigma\mapsto k_\sigma$. Given the preceding condition that $\ker(\tau)\subseteq \mathfrak A_n$, a necessary and sufficient condition for the inclusion $Z_g\subseteq \mathfrak A_n$ will be that the image of this section consists of even permutations. This section is a morphism of groups, so it suffices to understand the sign of $k_\sigma$ when $\sigma$ consists of a cycle $(c_1,\dots,c_s)$ in $\mathfrak S_{n_i}$ and is trivial on the other factors. Then $\ell(c_1)=\dots=\ell(c_s)$, by definition of $\sigma$. The formula $k_\sigma(g^n a_c)=g^n a_{\sigma(c)}$ shows that the non trivial cycles of $k_\sigma$ are of the form $(g^n a_{c_1},\dots, g^n a_{c_s})$; they all have the same length, $s$, and there are $\ell(c_1)$ of them. Consequently, the sign of $k_\sigma$ is equal to $(-1)^{(s-1)\ell(c_1)}=(-1)^{s-1}$ since $\ell(c_1)$ is odd. This proves that the sign of $k_\sigma$ is equal to the sign of $\sigma$. In addition to the condition that the orbits of $g$ have odd cardinalities, a necessary and sufficient condition for the image of $\sigma\mapsto k_\sigma$ to be contained in $\mathfrak A_n$ is thus that all symmetric groups $\mathfrak S_{n_i}$ coincide with their alternating groups, that is, $n_i\leq 1$ for all $i$. We can now conclude:

Theorem. — Let $1^{n_1}\dots r^{n_r}$ be a partition of $n$.

1. If $n_i=0$ for even $i$, and $n_i\leq 1$ for all $i$, then there exist two permutations in $\mathfrak S_n$ with orbit type $1^{n_1}\dots r^{n_r}$ which are not conjugate in $\mathfrak A_n$.
2. Otherwise, any two permutations in $\mathfrak S_n$ with that orbit type are conjugate in $\mathfrak A_n$.

We can check the initial example of two 5-cycles in $\mathfrak S_5$ which are not conjugate in $\mathfrak A_5$. Their orbit type is $5^1$: the only length that appears is 5, hence odd, and it has multiplicity $1$. In fact, this is the only orbit type in $\mathfrak S_5$ where this phenomenon appears!

Simple cases for non-simplicity

A subgroup $H$ of a group $G$ is said to be normal if one has $gH=Hg$ for every $g\in G$. The French terminology is that $H$ be distingué, and I must say that I find the English terminology definitely not illuminating, for it is quite a peculiar property for a subgroup $H$ to be normal. The definition itself my seem not illuminating and I start here with two better ways of introducing the notion of a normal subgroup.

1. An equivalence relation $\sim$ on $G$ is compatible with its group structure (namely, $g\sim g'$ and $k\sim k'$ imply $gk\sim g'k'$ for all $g,g',k,k'\in G$), so that the group structure of $G$ induces a group structure on the quotient set $G/\mathord\sim$ if and only if the class $H$ of the neutral element is a normal subgroup of $G$ and the relation satisfies: $g\sim g'$ if and only if $g'g^{-1}\in H$. 
2. Another definition could start with saying that $H$ is a normal subgroup if and only if there exists a morphism of groups $f\colon G\to G'$ such that $H=\ker(f)$. This is reminiscent of MacLane's definition of an abelian category (monomorphisms are kernels).

As I just said, being a normal subgroup is quite an extraordinary property, and a given group needs not having any other normal subgroup than the obvious ones: the full group, and the group reduced to the neutral element. These groups are called simple, with the exception of the trivial group which, by convention, is not said to be simple.

In a commutative group, every subgroup is normal; this is at least one useful point for the initial definition, to make this obvious. From this, one sees that a commutative group is simple if and only if it is cyclic of prime order.

Many exercises of undergraduate algebra ask to prove that a given subgroup of a given group is normal, or that a given group is not simple, and I discuss here a recent paper (Geller, Sue. « Simply Not a Simple Group ». The American Mathematical Monthly 127, n^o 4 (20 avril 2020): 352‑53. https://doi.org/10.1080/00029890.2020.1704167), starting with a classical exercise.

Let $G$ be a group and let $H$ be a subgroup of index $2$. Then $H$ is a normal subgroup.

Let $g\in G$. If $g\in H$, then $gH=H=Hg$, because $H$ is a subgroup. Otherwise, the group $G$ splits as the disjoint union of $H$ and $gH$, as well as the disjoint union of $H$ and $Hg$, because $H$ has index $2$. Consequently, $gH=G\setminus H=Hg$.

Let $G$ be a finite group and let $H$ be a subgroup whose index $p$ is the least prime factor of $\mathop{\mathrm{Card}}(G)$. Then $H$ is a normal subgroup.

Here I wish to let $G$ act on the left on the quotient set $G/H$. This action is transitive, and this furnishes a nontrivial morphism of groups from $G$ to the symmetric group $\mathfrak S(G/H)\simeq \mathfrak S_p$. Let $G'$ be the image of this morphism and let $N'$ be its cardinality. Since $N'$ divides the cardinality of $G$, the prime factors of $N'$ are at least $p$. However, $N'$ divides the cardinality of $\mathfrak S_p$, which is $p!$, so that the prime factors of $N'$ are at most $p$. Two cases are possible, either $N'=1$ (contradicting the fact that the initial action is transitive), or $N'=p$: the image of $f$ has cardinality $p$ so that $\ker(f)$ is a subgroup of index $p$. If $g\in\ker(f)$, one has in particular $gH=H$, so that $\ker(f)\subset H$. Since $H$ has also index $p$, we have $H=\ker(f)$, and $H$ is a normal subgroup.

Remark that if $H$ is a subgroup of a group $G$, and $G$ acts on the left on $G/H$, the kernel of the corresponding group morphism from $G$ to $\mathfrak S(G/H)$ is the intersection of all $gHg^{-1}$, the largest normal subgroup of $G$ which is contained in $H$.

A class of groups which are most often not simple are $p$-groups, that is, finite groups whose cardinality is a power of a prime number $p$. Indeed, such a group is never simple, unless it is a cyclic group of order $p$. This follows from the consideration of the action of $G$ on itself by conjugation. Assuming $G\neq 1$, the conjugacy classes of central elements have cardinality $1$, while the cardinality of the conjugacy class of a non-central element is divisible by $p$; the class equation implies that $\mathop{\mathrm{Card}}(G)-\mathop{\mathrm{Card}}(Z)$ is divisible by $p$, where $Z$ is the centrum of $G$; then $\mathop{\mathrm{Card}}(Z)$ is divisible by $p$, so that it is nontrivial, unless $Z=G$, that is, $G$ is commutative, and then cyclic of order $p$.

I can now discuss Sue Geller's note, which gives a criterion for non-simplicity of some finite groups, based on the existence of natural group morphisms from $G$ to permutation groups, given by the action of $G$ in the set of its Sylow subgroups.

Let $G$ be a finite group and let $p$ be a prime number. Write $n=\mathop{\mathrm{Card}}(G)=p^s m$, where $s\geq0$ and $p$ does not divide $n$. A subgroup $S$ of $G$ is a $p$-Sylow subgroup if its cardinality is equal to $p^s$. The Sylow theorems state the following properties of the set $\Sigma_p$ of $p$-Sylow subgroups and its cardinality $\sigma_p$.

1. The set $\Sigma_p$ is not empty (1st Sylow theorem); more generally, $G$ admits subgroups of all cardinalities dividing $p^s$.
2. One has $\sigma_p \equiv 1\pmod p$ and $\sigma_p\mid m$ (3rd Sylow theorem).
3. The map $g\mapsto gSg^{-1}$ induces an action of $G$ on the set $\Sigma_p$, and this action is transitive (2nd Sylow theorem), so that $\sigma_p=(G:N_G(S))$.
   

In particular, if $\sigma_p=1$, then the unique $p$-Sylow subgroup of $G$ is normal and $G$ is not simple (unless its order is a power of $p$, in which case $G$ is simple if and only if $G$ is cyclic of order $p$.

The book (I. Martin Isaacs, Finite Group Theory, Graduate Studies in Mathematics, v. 92 (Providence, R.I: American Mathematical Society, 2008) refines the congruence $\sigma_p\equiv 1\pmod p$. I give it here as a corollary.

Corollary. — Let $t$ be an integer such that for every distinct $p$-Sylow subgroups $S$ and $T$ of $G$, one has $\mathop{\mathrm{Card}}(S\cap T) \leq p^t$. Then $\sigma_p \equiv 1 \pmod{s-t}$.

Let $S$ be a $p$-Sylow subgroup of $G$ and let us restrict the action of $G$ on $\Sigma_p$ to n action of $S$. The orbit of $S$ is $S$ itself; its cardinality is $1$. If $T$ is another $p$-Sylow subgroup of $G$, then its orbit has cardinality $(S:Q)$, where $Q=S\cap N_G(T)$ is the set of all $s\in S$ such that $sTs^{-1}=T$. Then $Q$ is a $p$-group, hence is contained in a Sylow subgroup of $N_G(T)$, but $T$ is a Sylow subgroup of $N_G(T)$ which is normal (by construction of the normalizer), so that $T$ is the unique $p$-Sylow subgroup of $N_G(T)$; this implies that $Q\subset T$, hence $Q=S\cap T$. Consequently, $\mathop{\mathrm{Card}}(T)$ divides $p^t$ and the cardinality of the orbit of $T$ is a multiple of $p^{s-t}$. The corollary follows by recalling that the sum of the cardinalities of the orbits is equal to $\sigma_p$.

The theorem stated by Sue Geller is the following. It is in fact a variant of corollary 1.3 from the book of Isaacs.

Theorem. — Assume that $\sigma_p>1$ and that $n$ does not divide $\sigma_p!$. Then $G$ is not simple.

The transitive action of $G$ on the set $\Sigma_p$ furnishes a non-trivial morphism of groups $f\colon G\to\mathfrak S(\Sigma_p)$. In particular, $\ker(f)\neq G$. If $f$ were injective, then $G$ would be isomorphic to a subgroup of $\mathfrak S(\Sigma_p)\simeq \mathfrak S_p$, hence its order would divide $\sigma_p!$, a contradiction. Consequently, $\ker(f)$ is a non-trivial normal subgroup of $G$, and $G$ is not simple.

Example 1. — A group $G$ of order $48$ is not simple. One has $48=2^4\cdot 3$. Then $\sigma_2=1$ or $\sigma_2=3$. If $\sigma_2=1$, then $G$ is not simple, as we saw above. Otherwise, $\sigma_2=3$, and since $48$ does not divide $3!=6$, the theorem shows that $G$ is not simple.

Example 2. — Any simple group $G$ of order $60$ is  isomorphic to the alternating group $\mathfrak A_5$. The proof will consist in constructing a subgroup $H$ of $G$ of index $5$; then the action of $G$ on $G/H$ induces a non-trivial morphism from $G$ to $\mathfrak S(G/H)\simeq\mathfrak S_5$; since $G$ is simple, this morphism is injective. The composition with the signature morphism must be injective as well, so that its image is contained in the alternating subgroup. Since $G$ and $\mathfrak A_5$ both have cardinality $60$, $f$ is an isomorphism from $G$ to $\mathfrak A(G/H)\simeq\mathfrak A_5$.
This subgroup $H$ will be constructed as the normalizer of a subgroup of order $2$ ou $4$. From $60=2^2\cdot3\cdot5$, one gets $\sigma_2\in\{1,3,5,15\}$. 

The case $\sigma_2=1$ contradicts the assumption of simplicity for $G$, and the case $\sigma_2=3$ contradicts the theorem above, for $60$ does not divide $6$. 

If $\sigma_2=5$, let $S$ be a $2$-Sylow subgroup of $G$ and let $H=N_G(S)$ be its normalizer; one has $(G:H)=\sigma_2=5$.

Assume finally that $\sigma_2=15$.  By the above corollary, since $\sigma_2=15\not\equiv 1\pmod 4$, there exist distinct $p$-Sylow subgroups $S$ and $T$ of $G$ such that $P=S\cap T\neq\{1\}$. Let $H=N_G(P)$ be the normalizer of $P$. One has $H\subset S$ and $H\subset T$ (so says Isaac, without saying why), so that the inclusion $S\subset H$ is strict. Moreover, the inclusion $H\subset G$ is strict because $S$ is normal in $H$, but not in $G$. If $(G:H)\neq 5$, then $(G:H)=3$, and the action of $G$ on $\mathfrak S(G/H)$ furnishes an embedding of $G$ into $\mathfrak S_3$, which is absurd since $G$ has cardinality $60$ and $\mathfrak S_3$ has cardinality 6.

Happy New Year!

As was apparently first noticed by Noam Elkies, 2016 is the cardinality of the general linear group over the field with 7 elements, $G=\mathop{\rm GL}(2,\mathbf F_7)$. I was mentoring an 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...]