# 189-570A: Higher Algebra I

## Assignment 2: Corrections.

1. (a) Everyone did this question correctly. The congugacy classes in S5 correspond to cycle shapes, and hence are parametrised b partitions of 5. The cardinalities of the conugacy classes can be computed by direct counting arguments.

(b) Every conjugacy class corresponding to the cycle shape of an even permutation belongs to A5. There are four of these, corresponding to the partitions 1+1+1+1+1, 2+2+1, 3+1+1, and 5, of cardinalities 1, 15, 20, and 24 respectively. It turns out that the first three S5-conjugacy classes are also A5-conjugacy classes, but the fourth can't possibly be, since 24 does not divide the order of A5, which is 60. In fact, this conjugacy class is a disjoint union of two conjugacy classes in A5, each of size 12.

(c) Note that G=GL2(Fp) has cardniality p(p-1)2(p+1), by a previous assignment. The conjugacy classes in G are of four types:
(i) The central conjugacy classes, consisting of a scalar matrix. There are p-1 of these classes, each with one element, and of course their union is the center of G, which is isomorphic to Fp*.
(ii) The conjugacy classes of matrices with a single eigenvalue a in Fp*, but which are not fully diagonalizable. There are (p-1) such classes. The centralizer of the matrix (a0 1a) has cardinality p(p-1), so that the size of each conjugacy class is (p-1)(p+1) = p2-1.
(iii) The conjugacy classes of matrices with distinct eigenvalues in Fp*. There are (p-1)(p-2)/2 such classes, and each has cardinality p(p+1) since the centralizer of a diagonal matrix with distinct eigenvalues can be seen to be the group of all diagonal matrices, which has cardinality (p-1)2.
(iv) The conjugacy classes of matrices with irreducible characteristic polynomial. There are (p2-p)/2 such classes, corresponding to the possible pairs of eigenvalues which belong to Fp2, the field with p2 elements, but not to Fp and are invariant under conjugation. The centralizer of such a matrix can be seen, by a direct calculation, to be a group of cardinality (p-1)(p+1) so that each conjugacy class of this type has size p(p-1).

(d) Each conjugacy class in G=GL2(Fp) represented by an element of determinant 1 is a subset of H=SL2(Fp) which is preserved by conjugation in H and is hence a disjoint union of a collection of H-conjugacy classes. Here is what happens with each of the conjugacy classes of type (i)-(iv) mentionned in (c).
(i) There are only two central classes, corresponding to the eigenvalues 1 and -1. The center of H is of size 2.
(ii) There are two G-conjugacy classes of matrices with repreated eigenvalues belonging to H, namely, the ones attached to the eigenvalue 1 and -1. But each of these two G-conjugacy classes breaks up into a disjoint union of two H-conjugacy classes, of size (p-1)(p+1)/2. There are thus 4 classes of this type.
(iii) There are (p-3)/2 conjugacy classes of matrices with distinct eigenvalues a and a-1 in Fp. (One needs to exclude a=1 and a=-1 since then a=a-1 and the matrix in question is a scalar.) Each conjugacy class has cardinality p(p+1).
(iv) There are (p-1)/2 conjugacy classes of matrices with irreducible characteristic polynomial, corresponding to the possible pairs of eigenvalues which belong to Fp2, the field with p2 elements, do not belong to Fp, are invariant under conjugation and whose product is 1. The centralizer of such a matrix can be seen, by a direct calculation, to be a group of cardinality (p+1) so that each conjugacy class of this type has size p(p-1).

2. (a) The Sylow subgroup can be taken to be the group of all upper triangular matrices with 1's on the diagonal, as most of you correctly wote.

(b) This question seems to have been obscure to most people, and indeed there was a subtlety that I had missed in posing the question. What I wanted you to show was that if H is a Sylow p-subgrop, there is an increasing sequence of vector spaces ordered by inclusion
(*) V1 in V2 in ... in Vn= Fpn
such that Vj has dimension j (such a sequence is sometimes called a flag) and such that each Vj is preserved by the natural action of H. Conversely, any flag determines a Sylow p-subgroup. (To see this, use an induction argument, based on question 7... so I should have asked this question after question 7!) Now, the statement that all Sylow p-subgroups are conjugate is merely the statement that GL2(Fp) acts transitively on the set of all flags. (In fact, more is true: the linear group even acts transitively on all possible bases for Fpn.)

(c) The number of distinct Sylow p-subgroups is equal to the number of distinct flags as in (*). There are (pn-1)/(p-1) possible choices for v1, and once this choice is made, the flag is completely determined by choosing a flag in Vn/V1, a space of one dimension lower. From this observation, it follows by induction that the number of Sylow p-subgroups is equal to
(pn-1)(pn-1-1)...(p2-1)/(p-1)n-1.

3. Almost everyone did this corrrectly. As some of you also pointed out, I should have mentionned that the sets X1 and X2 are assumed to be finite: otherwise, (b) is false in general...

4. For this question, one had to show that if chiV(g) =n (resp -n), then rhoV(g) is the identity matrix (resp the negative of the identity matrix). This follows from the fact that chiV(g) is the sum of the n eigenvalues of rhoV(g), and since these eigenvalues are roots of unity (here, it is crucial that we assume G is finite!) it follows that the maximum (resp minium) for the trace is n (resp -n) with equality attained only for the identity matrix (resp its negative).

5. This was essentially done in class under the assumption that the characteristic of F does not divide the order of the group so that the matrices in the images of the varios representations are diagonalisable. To avoid this assumption, one could proceed by a direct calculation (which is what many of you did, correctly.)

6. (a) This was essentially done in class, where we saw that V decomposes as a direct sum of the one-dimensional space of constant functions, and the two-dimensional space of functions of sum 0.

(b) A direct calculation shows that (even in characteristic 3) the subspace of V of vectors which are fixed by the cyclic permutation g of order 3 is the space V0 of constant functions. Now suppose that V decomposes as a direct sum V0+W where W is two dimensional. By the result of part 7, the space W contains a vector which is fixed by g, so that the space of vectors in V fixed by G has dimension at least 2, contradicting the fact that this speace is one-dimensional.

7. Let V be such a representation. Choose a vector v in V and let W be the Fp vector space spanned by the G-translates of v. Since G is finite, W is a finite-dimensional Fp vector space, and it is preserved by G by construction. Now, decompose W into a disjoint union of G-orbits. Because W is a finite set, or cardinality a power of p, and because at least one orbit has size one (the orbit consisting of the 0 vector), there must be other orbits of size 1. These other orbits correspond to non-zero vectors in W which are fixed by all g in G. Let w be such a vector. The F-vector space spanned by w is a subrepresentation of V isomorphic to the trivial representation, and hence V, being irreducible, is itself the trivial representation.

8. (a) Almost everyone was able to show that if X1 and X2 are isomrphic as G-sets, then the associated permutation representations are isomorphic as well: this is a formal consequence of the functoriality of the assignment X --> Fct(X,F).

(b) Let G be the Dihedral group of order 8, viewed as he symmetry group of a square. Label the sides of the square as A,B,C,D, and the diagonals as E and F. Label the vertices of the square as 1,2,3,4 and the segments joining the midpoints of opposite sides as 5 and 6. Now let X1={A,B,C,D,E,F} and X2={1,2,3,4,5,6} with the natural G actions. Then X1 and X2 are not isomorphic as G-sets, as was explained in class, even though the characters of the associated permutation representations are the same and hence these representations are isomorphic.

Note that X1 and X2 are not transitive; but transitive counterexamples exist as well. After doing assignment 3, you might want to ponder the following one: let G=GL3(F2) acting on V=F23 in the natural way. Let X1 be the set of non-zero vectors in V, and let X2 be the set of two-dimensional subspaces of V. Both X1 and X2 are transitive G-sets of cardinality 7, which are not isomorphic as G-sets. However, their associated permutation representations are isomorphic...