Math 5125
Monday, September 12
Third Homework Solutions
1. The action of
G
on
G
/
H
(the left cosets of
H
in
G
) yields a homomorphism
θ
:
G
→
S
n
, and by hypothesis
θ
(
G
)
⊆
A
n
. Note that
θ
(
g
)
has order 1 or 2, because
g
has order
2. Since
θ
(
g
)
∈
A
n
and
n
4, we see that
θ
(
g
)
must fix an element of
A
n
(ungraded
homework of August 30). Now the permutation
θ
(
g
)
is determined by labeling the left
cosets of
H
1
,
2
,...,
n
, and sending the coset
yH
to
gyH
. This means that
θ
(
g
)
must
fix one of the left cosets
yH
; in other words, there exists
x
∈
G
such that
gxH
=
xH
.
2. The elements of order 2 in
S
6
have cycle shape (1 2), (1 2)(3 4), (1 2)(3 4)(5 6). The
sizes of the conjugacy classes are 6
*
5
/
2
=
15, 6
*
5
*
4
*
3
/
(
2
*
2
*
2
) =
45 and
6!
/
(
2
*
2
*
2
*
3!
) =
15. Since

S
6

=
720, the sizes of the centralizers are

C
G
(
1 2
)

=
720
/
15
=
48,

C
G
((
1 2
)(
3 4
))

=
720
/
45
=
16,

C
G
((
1 2
)(
3 4
)(
5 6
))

=
48.
3. The sizes of the conjugacy classes divide 21, so must be 1,3,7 or 21. Since Z
(
G
) =
1,
there is 1 conjugacy class of size 1; also there is no conjugacy class of size 21. Let
x
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 Fall '07
 PALinnell
 Algebra, Sets, Coset, Conjugacy class, Index of a subgroup, conjugacy classes, left cosets

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