21.
Permutation groups II
21.1.
Conjugacy classes.
Let
G
be a group, and consider the following
relation
∼
on
G
: given
f, h
∈
G
, we put
f
∼
h
⇐⇒
there exists
g
∈
G
s.t.
h
=
gfg

1
.
Thus, in the terminolgy from Lecture 20,
f
∼
h
⇐⇒
h
is a conjugate of
f
.
Definition.
The relation
∼
is called the conjugacy
relation.
Lemma 21.0.
The conjugacy relation
∼
is an equivalence relation on
G
.
Proof.
(i) Reflexivity: for every
f
∈
G
we have
efe

1
=
f
, so
f
∼
f
(that
is,
f
is a conjugate of itself).
(ii) Symmetry: assume that
f
∼
h
, so that
h
=
gfg

1
for some
g
∈
G
. Then
f
=
g

1
hg
=
uhu

1
where
u
=
g

1
∈
G
. Therefore,
h
∼
f
.
(iii) Transitivity: assume that
f
∼
h
and
h
∼
k
, so that
h
=
g
1
fg

1
1
and
k
=
g
2
hg

1
2
for some
g
1
, g
2
∈
G
.
Then
k
=
g
2
g
1
fg

1
1
g

1
2
=
gfg

1
where
g
=
g
2
g
1
∈
G
, so
f
∼
k
.
Having established that
∼
is symmetric, we can safely use the terminology
“
f
and
h
are conjugate” instead of saying “
h
is a conjugate of
f
.”
Definition.
The equivalence classes with respect to the conjugacy relation
are called the conjugacy classes
of
G
. For each
f
∈
G
we denote its conjugacy
class by
K
(
f
). Thus,
K
(
f
) =
{
h
∈
G
:
h
=
gfg

1
for some
g
∈
G
}
.
Note that by the general properties of equivalence classes, conjugacy classes
form a partition of
G
, that is, distinct conjugacy classes are disjoint, and
the union of all conjugacy classes of
G
is the entire group
G
.
Warning:
Conjugacy classes should not be confused with cosets.
21.2.
Conjugacy classes in
S
n
.
Computation of conjugacy classes in a
given group may be a complicated problem. However, conjugacy classes in
permutation groups admit a very simple and explicit description.
We start with an example showing how conjugation works in
S
n
.
1
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Example:
Let
n
≥
3, let
f
= (1
,
2
,
3)
∈
S
n
, and let
g
be some element of
S
n
. Let us compute the conjugate
gfg

1
=
g
(1
,
2
,
3)
g

1
.
To do this we have to track the image of each
i
∈ {
1
, . . . , n
}
under the
composed map.
First we analyze where
gfg

1
sends
g
(1),
g
(2) and
g
(3).
We have
g
(1)
g

1
→
1
f
→
2
g
→
g
(2)
g
(2)
g

1
→
2
f
→
3
g
→
g
(3)
g
(3)
g

1
→
3
f
→
1
g
→
g
(1)
.
Now take any
i
=
g
(1)
, g
(2) or
g
(3).
Then (since
g
is bijective) we have
g

1
(
i
) = 1
,
2 or 3, and therefore
f
(
g

1
(
i
)) =
g

1
(
i
). Thus we get
i
g

1
→
g

1
(
i
)
f
→
g

1
(
i
)
g
→
i.
So,
gfg

1
maps
g
(1) to
g
(2),
g
(2) to
g
(3),
g
(3) to
g
(1), and fixes all
other elements. Therefore,
gfg

1
=
g
(1
,
2
,
3)
g

1
= (
g
(1)
, g
(2)
, g
(3)).
It is not hard to see that similar formula is true in general: for any cycle
(
i
1
, . . . , i
k
) and any
g
∈
S
n
we have
g
(
i
1
, . . . , i
k
)
g

1
= (
g
(
i
1
)
, . . . , g
(
i
k
))
.
(K1)
In other words, if
f
is a cycle of length
k
, then
gfg

1
is also a cycle of length
k
whose entries are obtained by applying
g
to the entries of
f
.
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 Winter '11
 PhanThuongCang
 Group Theory, k2, Coset, Index of a subgroup, conjugacy classes

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