6Symmetric groups II

IA Groups



6.2 Conjugacy classes in A
n
We have seen that
|S
n
|
= 2
|A
n
|
and that conjugacy classes in
S
n
are “nice”.
How about in A
n
?
The first thought is that we write it down:
ccl
S
n
(σ) = {τ S
n
: (ρ S
n
) τ = ρσρ
1
}
ccl
A
n
(σ) = {τ A
n
: (ρ A
n
) τ = ρσρ
1
}
Obviously
ccl
A
n
(
σ
)
ccl
S
n
(
σ
), but the converse need not be true since the
conjugation need to map σ to τ may be odd.
Example.
Consider (1 2 3) and (1 3 2). They are conjugate in
S
3
by (2 3), but
(2 3)
6∈ A
3
. (This does not automatically entail that they are not conjugate in
A
3
because there might be another even permutation that conjugate (1 2 3) and
(1 3 2). In A
5
, (2 3)(4 5) works (but not in A
3
))
We can use the orbit-stabilizer theorem:
|S
n
| = | ccl
S
n
(σ)||C
S
n
(σ)|
|A
n
| = | ccl
A
n
(σ)||C
A
n
(σ)|
We know that
A
n
is half of
S
n
and
ccl
A
n
is contained in
ccl
S
n
. So we have two
options: either
ccl
S
n
(
σ
) =
ccl
A
n
(
σ
) and
|C
S
n
(
σ
)
|
=
1
2
|C
A
n
(
σ
)
|
; or
1
2
| ccl
S
n
(
σ
)
|
=
| ccl
A
n
(σ)| and C
A
n
(σ) = C
S
n
(σ).
Definition
(Splitting of conjugacy classes)
.
When
| ccl
A
n
(
σ
)
|
=
1
2
| ccl
S
n
(
σ
)
|
, we
say that the conjugacy class of σ splits in A
n
.
So the conjugacy classes are either retained or split.
Proposition.
For
σ A
n
, the conjugacy class of
σ
splits in
A
n
if and only if
no odd permutation commutes with σ.
Proof.
We have the conjugacy classes splitting if and only if the centralizer
does not. So instead we check whether the centralizer splits. Clearly
C
A
n
(
σ
) =
C
S
n
(
σ
)
A
n
. So splitting of centralizer occurs if and only if an odd permutation
commutes with σ.
Example. Conjugacy classes in A
4
:
Cycle type Example | ccl
S
4
| Odd element in C
S
4
? | ccl
A
4
|
(1, 1, 1, 1) e 1 Yes (1 2) 1
(2, 2) (1 2)(3 4) 3 Yes (1 2) 3
(3, 1) (1 2 3) 8 No 4, 4
In the (3, 1) case, by the orbit stabilizer theorem,
|C
S
4
((1 2 3))
|
= 3, which is
odd and cannot split.
Example. Conjugacy classes in A
5
:
Cycle type Example | ccl
S
5
| Odd element in C
S
5
? | ccl
A
5
|
(1, 1, 1, 1, 1) e 1 Yes (1 2) 1
(2, 2, 1) (1 2)(3 4) 15 Yes (1 2) 15
(3, 1, 1) (1 2 3) 20 Yes (4 5) 20
(5) (1 2 3 4 5) 24 No 12, 12
Since the centralizer of (1 2 3 4 5) has size 5, it cannot split, so its conjugacy
class must split.
Lemma. σ = (1 2 3 4 5) S
5
has C
S
5
(σ) = hσi.
Proof. | ccl
S
n
(
σ
)
|
= 24 and
|S
5
|
= 120. So
|C
S
5
(
σ
)
|
= 5. Clearly
hσi C
S
5
(
σ
).
Since they both have size 5, we know that C
S
5
(σ) = hσi
Theorem. A
5
is simple.
Proof.
We know that normal subgroups must be unions of the conjugacy classes,
must contain
e
and their order must divide 60. The possible orders are 1, 2,
3, 4, 5, 6, 10, 12, 15, 20, 30. However, the conjugacy classes 1, 15, 20, 12, 12
cannot add up to any of the possible orders apart from 1 and 60. So we only
have trivial normal subgroups.
In fact, all
A
n
for
n
5 are simple, but the proof is horrible (cf. IB Groups,
Rings and Modules).