As was apparently first noticed by Noam Elkies, 2016 is the cardinality of the general linear group over the field with 7 elements, . 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 . The first column of a matrix in must be non-zero, hence there are possibilities; for the second column, it only needs to be non-collinear to the first one, and each choice of the first column forbids second columns, hence possibilities. In the end, one has . The same argument shows that the cardinality of the group is equal to .
Let's go back to our example. The factorization of this cardinal comes easily: . Consequently, there are three Sylow subgroups to find, for the prime numbers , and .
The cas 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 . What makes things work is that is the characteristic of the chosen finite field. In general, if is a power of , then the subgroup of upper-triangular matrices in with s one the diagonal has cardinality , which is exactly the highest power of divising the cardinality of .
Let's now study . We need to find a group of order inside . There are a priori two possibilities, either , or .
We will find a group of the first sort, which will that the second case doesn't happen, because all -Sylows are pairwise conjugate, hence isomorphic.
Now, the multiplicative group is of order , and is cyclic, hence contains a subgroup of order , namely . Consequently, the group of diagonal matrices with coefficients in is isomorphic to and is our desired -Sylow.
Another reason why does not contain a subgroup isomorphic to is that it does not contain elements of order . Let's argue by contradiction and consider a matrix such that ; then its minimal polynomial divides . Since , the matrix is diagonalizable over the algebraic closure of . The eigenvalues of are eigenvalues are th roots of unity, and are quadratic over since . On the other hand, if is a th root of unity belonging to , one has , hence since . Consequently, is a cubic root of unity and , showing that has order .
It remains to treat the case , which I find slightly trickier. Let's try to find elements in whose order divides . As above, it is diagonalizable in an algebraic closure, its minimal polynomial divides , and its roots belong to , hence satisfy , hence . Conversely, is cyclic of order , hence contains an element of order , and such an element is quadratic over , hence its minimal polynomial has degree . The corresponding companion matrix in is an element of order , generating a subgroup of isomorphic to . We also observe that (because its square is ); since is diagonalizable in an algebraic closure with as the only eigenvalue, this shows .
Now, there exists a -Sylow subgroup containing , and will be a normal subgroup of (because its index is the smallest prime number dividing the order of , which is ). This suggests to introduce the normalizer of in . One then has . Let be such that ; then there exists a unique such that , and (because has order modulo ), hence —in other words, .
There exists a natural choice of : the involution () which exchanges the two eigenspaces of . To finish the computation, it's useful to take a specific example of polynomial of degree whose roots in are primitive th roots of unity. In other words, we need to factor the th cyclotomic polynomial over and find a factor of degree ; actually, Galois theory shows that all factors have the same degree, so that there should be 4 factors of degree . To explain the following computation, some remark is useful. Let
be a th root of unity in ; we have
but , hence . If
is the minimal polynomial of , the other root is ,
hence the constant term of is equal to .
We start from and observe that , so that . To find the factors of degree , we remember that their constant terms should be equal to . We thus go on differently, writing and solving for : this gives , hence and . The other factors are found similarly and we get
We thus choose the factor and set .
Two eigenvectors for are and , where is the other root of . The equations for are and ; this gives . The subgroup generated by and has order and is a -Sylow subgroup of .
Generalizing this method involves finding large commutative -subgroups (such as ) which belong to appropriate (possibly non-split) tori of and combining them with adequate parts of their normalizer, which is close to considering Sylow subgroups of the symmetric group. The paper Sylow -subgroups of the classical groups over finite fields with characteristic prime to 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 -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...]
Tuesday, February 9, 2016
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