**189-570A:** Higher Algebra I

## Assignment 2: Corrections.

**1**. (a) Everyone did this question correctly. The congugacy classes in
S_{5} 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 A_{5}. 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 S_{5}-conjugacy classes are also
A_{5}-conjugacy classes, but the fourth can't possibly be, since 24
does not divide the order of A_{5}, which is 60.
In fact, this conjugacy class is a disjoint union of two
conjugacy classes in A_{5}, each of size 12.

(c) Note that G=GL_{2}(F_{p}) 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 F_{p}^{*}.

(ii) The conjugacy classes of matrices with a single eigenvalue
a in F_{p}^{*}, but which are not fully diagonalizable.
There are (p-1) such classes.
The centralizer of the matrix (^{a}_{0}^{ 1}_{a})
has cardinality p(p-1), so that the size of each conjugacy class is
(p-1)(p+1) = p^{2}-1.

(iii) The conjugacy classes of matrices with distinct eigenvalues in
F_{p}^{*}. 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 (p^{2}-p)/2 such classes, corresponding to the
possible pairs of eigenvalues which belong to F_{p2},
the field with p^{2} elements, but not to F_{p} 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=GL_{2}(F_{p}) represented by an
element of determinant 1 is a subset of H=SL_{2}(F_{p})
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 F_{p}. (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 F_{p2},
the field with p^{2} elements, do not belong to F_{p},
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

(*) V_{1} in V_{2} in ... in V_{n}= F_{p}^{n}

such that V_{j} has dimension j (such a sequence is sometimes called a
*flag*) and such that each V_{j} 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
GL_{2}(F_{p}) acts transitively on the set of all
flags.
(In fact, more is true: the linear group even acts transitively on all
possible bases for F_{p}^{n}.)

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

(p^{n}-1)(p^{n-1}-1)...(p^{2}-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 X_{1} and X_{2} are
assumed to be
finite: otherwise, (b) is false in general...

**4**. For this question, one had to show that if chi_{V}(g)
=n (resp -n), then rho_{V}(g) is the identity matrix
(resp the negative of the identity matrix).
This follows from the fact that chi_{V}(g) is
the sum of the n eigenvalues of
rho_{V}(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 V_{0}
of constant functions. Now suppose that V decomposes
as a direct sum V_{0}+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 F_{p} vector space spanned by the G-translates of v.
Since G is finite, W is a finite-dimensional F_{p} 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 X_{1} and X_{2} 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 X_{1}={A,B,C,D,E,F} and
X_{2}={1,2,3,4,5,6} with the natural G actions.
Then X_{1} and X_{2} 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 X_{1} and X_{2} are not transitive; but transitive counterexamples exist as well.
After doing assignment 3, you might want to ponder the following one:
let G=GL_{3}(F_{2}) acting on
V=F_{2}^{3} in
the natural way. Let X_{1} be the set of non-zero vectors
in V, and let X_{2} be the set of two-dimensional subspaces of V.
Both X_{1} and X_{2} are transitive
G-sets of cardinality 7, which are not isomorphic as G-sets.
However, their associated permutation representations are isomorphic...