6Symmetric groups II

IA Groups 6.1 Conjugacy classes in S
n
Recall σ, τ S
n
are conjugate if ρ S
n
such that ρσρ
1
= τ.
We first investigate the special case, when σ is a k-cycle.
Proposition.
If (
a
1
a
2
· · · a
k
) is a
k
-cycle and
ρ S
n
, then
ρ
(
a
1
· · · a
k
)
ρ
1
is the k-cycle (ρ(a
1
) ρ(a
2
) · · · ρ(a
3
)).
Proof.
Consider any
ρ
(
a
1
) acted on by
ρ
(
a
1
· · · a
k
)
ρ
1
. The three permutations
send it to
ρ
(
a
1
)
7→ a
1
7→ a
2
7→ ρ
(
a
2
) and similarly for other
a
i
s. Since
ρ
is
bijective, any
b
can be written as
ρ
(
a
) for some
a
. So the result is the
k
-cycle
(ρ(a
1
) ρ(a
2
) · · · ρ(a
3
)).
Corollary.
Two elements in
S
n
are conjugate iff they have the same cycle type.
Proof.
Suppose
σ
=
σ
1
σ
2
· · · σ
`
, where
σ
i
are disjoint cycles. Then
ρσρ
1
=
ρσ
1
ρ
1
ρσ
2
ρ
1
· · · ρσ
`
ρ
1
. Since the conjugation of a cycle conserves its length,
ρσρ
1
has the same cycle type.
Conversely, if σ, τ have the same cycle type, say
σ = (a
1
a
2
· · · a
k
)(a
k+1
· · · a
k+`
), τ = (b
1
b
2
· · · b
k
)(b
k+1
· · · b
k+`
),
if we let ρ(a
i
) = b
i
, then ρσρ
1
= τ.
Example. Conjugacy classes of S
4
:
Cycle type Example element Size of ccl Size of centralizer Sign
(1, 1, 1, 1) e 1 24 +1
(2, 1, 1) (1 2) 6 4 1
(2, 2) (1 2)(3 4) 3 8 +1
(3, 1) (1 2 3) 8 3 +1
(4) (1 2 3 4) 6 4 1
We know that a normal subgroup is a union of conjugacy classes. We can now
find all normal subgroups by finding possible union of conjugacy classes whose
cardinality divides 24. Note that the normal subgroup must contain e.
(i) Order 1: {e}
(ii) Order 2: None
(iii) Order 3: None
(iv)
Order 4:
{e,
(1 2)(3 4)
,
(1 3)(2 4)
,
(1 4)(2 3)
}
=
C
2
× C
2
=
V
4
is a possible
candidate. We can check the group axioms and find that it is really a
subgroup
(v) Order 6: None
(vi) Order 8: None
(vii)
Order 12:
A
4
(We know it is a normal subgroup since it is the kernel of
the signature and/or it has index 2)
(viii) Order 24: S
4
We can also obtain the quotients of
S
4
:
S
4
/{e}
=
S
4
,
S
4
/V
4
=
S
3
=
D
6
,
S
4
/A
4
=
C
2
, S
4
/S
4
= {e}.