MATH 594, WINTER 2006, PROBLEM SET 1
DUE: 1/20/2006
Exercises to be handed in
Exercise 1. Make a table of all conjugacy classes of S6 . For each conjugacy class,
compute its cardinality, and the cardinality of the centralizer of a representant.
(Check: Do the cardinalities of the conjugacy classes add up to 6! = 720?)
Solution.
example
card. of conj. class card. of centralizer
partition
1, 1, 1, 1, 1, 1
1
1
720
6·5
(1 2)
= 15
720/15 = 48
2, 1, 1, 1, 1
2
4·3
(1 2)(3 4)
( 6·5
720/45 = 16
2, 2, 1, 1
·
)/2
=
45
2
2
6·5 4·3 2·1
2, 2, 2
(1 2)(3 4)(5 6) ( 2 · 2 · 2 )/3! = 15
720/15 = 48
6·5·4
(1 2 3)
720/40 = 18
3, 1, 1, 1
=
40
3
6·5·4 3·2
3, 2, 1
· 2 = 120
(1 2 3)(4 5)
720/120 = 6.
3
3·2·1
(1 2 3)(4 5 6)
( 6·5·4
·
)/2
=
40
720/40 = 18
3, 3
3
3
6·5·4·3
4, 1, 1
(1 2 3 4)
= 90
720/90 = 8
4
6·5·4·3 2·1
(1 2 3 4)(5 6)
·
=
90
720/90 = 8
4, 2
4
2
6·5·4·3·2
(1 2 3 4 5)
= 144
720/144 = 5
5, 1
5
6·5·4·3·2·1
6
(1 2 3 4 5 6)
=
120
720/120 = 6
6
Exercise 2. [DF], §4.1, 7, Exercise 7. Here you will learn the notion of a
primitive permutation representation. A group is called primitive if it has a
faithful primitive permutation representation.
Solution.
(a) Clearly 1(B) = B so 1 ∈ GB . If σ, τ ∈ GB then στ (B) = σ(τ (B)) =
σ(B) = B, so στ ∈ GB . Finally if σ ∈ GB , then σ −1 (B) = σ −1 (σ(B)) =
1(B) = B, so σ −1 ∈ GB . This shows that GB contains 1 and is closed
under multiplication and inverse. We conclude that GB is a group.
Suppose that σ ∈ Ga and a ∈ B. Then a = σ(a) ∈ σ(B) and a ∈
σ(B) ∩ B. So σ(B) ∩ B 6= ∅ and σ(B) = B by the definition of a block.
So σ ∈ GB . We have shown that Ga ⊆ GB .
(b) Suppose that σi (B) ∩ σj (B) 6= ∅. Then
σi−1 (σi (B) ∩ σj (B)) = B ∩ (σi−1 σj )(B).
1
2
DUE: 1/20/2006
Either B = σi−1 σj (B) and σi (B) = σj (B) and i = j or B ∩ σi−1 σj (B) = ∅
and
σi (B) ∩ σj (B) = σi (B ∩ σi−1 σj (B)) = σi (∅) = ∅.
This shows that σ1 (B), . . . , σn (B) are disjoint. Let a ∈ B. If b ∈ A then
there exists a σ ∈ G such that σ(a) = b. It follows that b ∈ σ(B). Since
σ(B) = σi (B) for some i we see that σ1 (B), . . . , σn (B) is a partition of A.
(c) Consider G = S4 acting on {1, 2, 3, 4}. A nontrivial block B has size 2.
(From part (b) follows that the size of a block is a divisor of |A|.) If B
is a block, then so is σ(B). We can choose σ such that σ(B) = {1, 2}.
If we take τ = (2 3) then τ ({1, 2}) = {2, 3}. But {1, 2} ∩ τ ({1, 2}) =
{1, 2} ∩ {2, 3} = {2} is nonempty and {1, 2} =
6 τ ({1, 2}) = {2, 3}. This
shows that {1, 2} is not a block. So the action of G = S4 is not primitive.
The group D8 acts on the 4 vertices of a square. If x, y are opposite
vertices of the square, then so are σ(x), σ(y) for any σ ∈ D8 . If we
order the vertices of the square by 1, 2, 3, 4 by going around the square in
counterclockwise fashion, then {1, 3} and {2, 4} are nontrivial blocks. So
the action of D8 on the vertices is not primitive.
(d) Suppose that the action is not primitive. Let B be a nontrivial block (so
B has more than 1 element and B 6= A). Let GB be its stabilizer. The
action of G on A is transitive, and the action of GB on A is not transitive
because the orbit of any a ∈ B is contained in B which is a strict subset of
A. This shows that GB 6= G. We have seen that GB contains Ga for any
a ∈ GB . The group GB acts on B. We claim that this action is transitive:
Suppose that a, b ∈ B. There exists an σ ∈ G such that σ(a) = b. But
then b ∈ σ(B) ∩ B, so B ∩ σ(B) 6= ∅ and we must have B = σ(B) and
σ ∈ GB . The group Ga does not act transitively on B because the orbit
of a ∈ B is just {a}. This shows that Ga 6= GB . So GB is a group strictly
in between Ga and G.
Suppose now that H is a strict subgroup of G, strictly containing Ga .
Define B = Ha as the H-orbit of a. Clearly B has more than 1 element
because H 6= Ga . Suppose that H acts transitively on A. Let σ ∈ G.
Then there exists a τ ∈ H such that τ σ(a) = a. So τ σ ∈ Ga ⊆ H
and therefore σ ∈ τ −1 H = H. This shows that G = H. Contradiction.
Hence H does not act transitively. Therefore B = Ha 6= A. We claim
that B is a block. Indeed, suppose that σ(B) ∩ B 6= ∅ for some σ ∈ G.
Say, b ∈ σ(B) ∩ B. We can write b = τ1 (a) and b = σ(τ2 (a)) for some
τ1 , τ2 ∈ H. So τ1−1 στ2 (a) = a and τ1−1 στ2 ∈ Ga ⊆ H. It follows that
σ ∈ H and σ(B) = σHa = Ha = B. So B is a nontrivial block for
the action of G. We conclude that the permutation representation is not
primitive.
Exercise 3. [DF], §4.1, Exercise 9.
PROBLEM SET 1
3
(a) We can write Oi = H · ai for some ai ∈ A. Suppose that g ∈ G. There
exists a j such that gai ∈ Oj = Haj . It follows that gOi = gHai =
Hgai = Haj = Oj . Since G acts transitively on A, it acts transitively
on {O1 , . . . , Or }. In particular for every i 6= j there exists a g such that
gOi = Oj . It follows that |Oj | = |gOi | = |Oi |, so all orbits have the same
cardinality.
(b) If a ∈ O1 then O1 = H · a. The group H acts transitively on O1 . So O1
is the H-orbit of a. The stabilizer of a (in H) is Ha = Ga ∩ H. We have
a natural bijection
H/Ha ∼
= H · a = O1 .
From this follows that |O1 | = |H : H ∩ Ga |. We have
r=
|A|
|G|/|Ga |
|G|/|Ga |
|G|
=
=
=
= |G : HGa |.
|O1 |
|H|/|H ∩ Ga |
|Ga H|/|Ga |
|Ga H|
Exercise 4. The group Sn acts on {1, 2, . . . , n} as usual. Let H be the stabilizer
of n (so H is isomorphic to Sn−1 ). Describe the set of double cosets
H\Sn /H = {HσH | σ ∈ Sn }
What is the cardinality of each double coset?
Solution. Double cosets in H\Sn /H are in one-to-one correspondence with
H-orbits in Sn /H. There is a natural bijection ψ : Sn /H ∼
= {1, 2, . . . , n} that
respects the action of Sn (Sn acts on Sn /H by left multiplication). Here H is the
stabilizer of n and H ∼
= S{1,2,...,n−1} = Sn−1 . So ψ(H) = n. Also ψ((n − 1 n)H) =
n − 1. There are clearly two H-orbits in {1, 2, . . . , n} namely {1, 2, . . . , n − 1}
and {n}. These two orbits correspond with the double cosets H(n − 1 n)H and
H1H = H. Obviously |H1H| = |H| = (n − 1)!. H(n − 1 n)H is the union of
n − 1 left cosets of H, so it has (n − 1)(n − 1)! elements.
Exercise 5. Suppose that G is a finite group and H is a proper subgroup. Let
K1 , K2 , . . . , Kr be the distinct conjugacy classes of H.
(a) Show that each Ki is contained in a unique conjugacy class of G (call it
Li ).
Solution. suppose that Ki is the conjugacy class of gi in H. Let Li be
the conjugacy class of gi in G. Clearly, if elements are conjugate in H
there are conjugate in G, so Ki ⊆ Li .
(b) Prove that
|Li | ≤ |G : H||Ki |
where |G : H| = |G|/|H| is the index of H in G.
Solution. Suppose that
G/H = {u1 H, u2 H, . . . , ur H}
4
DUE: 1/20/2006
were r = |G : H|. If l ∈ Li then we can write l = kgi k −1 for some
k ∈ G. Write k = uj h for some h ∈ H. then l = kgi k −1 = uj hgi h−1 u−1
j ∈
uj Ki u−1
.
This
shows
that
j
Li ⊆
r
[
ui Ki u−1
i .
i=1
So we get
|Li | ≤ r|Li | = |G : H||Li |.
(c) Suppose that L1 , L2 , . . . , Lr are all conjugacy classes of G. Prove that
|Li | = |G : H||Ki | for all i. Derive a contradiction by looking at the
conjugacy class of 1.
Solution. Suppose that L1 , . . . , Lr are all conjugacy classes (but some
conjugacy classes may appear more than once). Choose indices i1 , i2 , . . . , is
such that Li1 , . . . , Lis are exactly all conjugacy classes, without repetition.
Then
|G| = |Li1 | + |Li2 | + · · · + |Lis | ≤ |G : H|(|Ki1 | + |Ki2 | + · · · + |Kis |) ≤
≤ |G : H|(|K1 | + · · · + |Kr |) = |G : H||H| = |G|.
So all the inequalities above are equalities. It follows that {i1 , . . . , is } =
{1, 2, . . . , r} and r = s. Moreover, we must have |Lij | = |G : H||Kij | for
all j, so |Li | = |G : H||Ki | for all i. For some i, Li is the conjugacy class
of 1, and Ki must be the conjugacy class of 1 in H. But then 1 = |Li | =
|G : H||Ki | = |G : H|, so |G : H| = 1 and H = G. Contradiction!
We have now proven the following theorem:
Theorem: If G is a finite group and H is a proper subgroup, then there exists a
conjugacy class L of G such that L ∩ H = ∅.
Hard Exercises (optional, for extra credit)
Exercise 6. [DF], §4.2, Exercise 8. (Not really hard, but you need the right
idea.)
Solution. Consider the action of G on G/H. Let K be the kernel of this action.
Then G/K acts faithfully on G/H. So G/K
of SG/H ∼
= Sn . So
T is a subgroup
−1
|G : K| = |G/K| ≤ |Sn | = n!. (Note: K = g∈G gHg .)
Exercise 7. [DF], §4.3, Exercise 26.
Solution. Define S = {(g, a) ∈ G × A | g · a = a}. For any a ∈ A, A ∼
= G/Ga ,
so |Ga | = |G|/|A|. Since
[
S=
Ga × {a}
a∈A
PROBLEM SET 1
we have
|S| =
X
|Ga | = |A|
a∈A
5
|G|
= |G|.
|A|
For g ∈ G, define Fix(g) = {a ∈ A | g · a = a}. We have
[
S=
{g} × Fix(g)
g∈G
so
|G| = |S| =
X
| Fix(g)|.
g∈G
So “on average”, Fix(g) has 1-element. Note that Fix(1) = |A| > 1. So
X
| Fix(g)| < |G| − 1.
g∈G\{1}
By the pigeonhole principle, there must exist a g ∈ G\{1} such that | Fix(g)| = 0,
i.e., Fix(g) = ∅.
Exercise 8. [DF], §4.3, Exercise 27. (The only proof that I can see uses Exercise 5.)
Solution. Let H be the group generated by g1 , . . . , gr . This group is abelian.
Every conjugacy class of G intersects with H. From Exercise 5 follows that
H = G. Thus, G is abelian.
Harm Derksen, 3067EH, 763 2309
Office hours: MW 1-2pm, F 3-4pm.
http://www.math.lsa.umich.edu/~hderksen/math594.w06/index.html