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Practice Problem Solutions
1.
Despite its appearance, this is a problem dealing exclusively with cosets and not using
Lagrange’s Theorem. We start with the following observation. Let
a, b
∈
K
and suppose
that
a
(
H
∩
K
) =
b
(
H
∩
K
). Then
a

1
b
∈
H
∩
K
≤
H
and so
aH
=
bH
. It follows that the
map
{
a
(
H
∩
K
)

a
∈
K
} → {
aH

a
∈
H
∨
K
}
a
(
H
∩
K
)
7→
aH
is welldeﬁned, i.e. does not depend on the choice of coset representative
a
. Moreover, this
map is onetoone. For if
a, b
∈
K
and
aH
=
bH
then
a

1
b
∈
H
and since
a

1
b
∈
K
we have
a

1
b
∈
H
∩
K
so that
a
(
H
∩
K
) =
b
(
H
∩
K
). It follows that the number of cosets in the set
on the left above (which is [
K
:
H
∩
K
]) is less than or equal to the number of cosets in the
set on the right above (which is [
H
∨
K
:
H
]). That is
[
K
:
H
∩
K
]
≤
[
H
∨
K
:
H
]
.
2.
In a recent homework assignment, we showed that the subgroup
G
=
{
σ
∈
S
4

σ
(4) = 4
}
is isomorphic to
S
3
. There’s nothing particularly special about the integer 4, and the same
technique can be used to show that the three subgroups
{
σ
∈
S
4

σ
(
i
) =
i
}
,
i
= 1
,
2
,
3 are all also isomorphic to
S
3
.
3.
Let (
ab
) be a transposition in
S
n
. If
a
= 1 or
b
= 1 then (
ab
) is already among
(12)
,
(13)
, . . . ,
(1
n
). So suppose that
a, b
6
= 1. Then, since
a
6
=
b
, it is trivial to verify that
(
ab
) = (1
a
)(1
b
)(1
a
)
.
It follows that any transposition can be written using only the transpositions (12)
,
(13)
, . . . ,
(1
n
),
and since any permutation can be written as a product of transpositions, that any element
of
S
n
can be written using only the transpositions (12)
,
(13)
, . . . ,
(1
n
).
4.
(a) Suppose that
σ
∈
S
n
is a 3cycle. Then
σ
has order 3. If
σ
6∈
H
then, since
σ
= (
σ
2
)
2
,
we must have
σ
2
6∈
H
as well. That is,
σH
6
=
H
and
σ
2
H
6
=
H
. But
H
has index 2 and so
only has two cosets in
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 Fall '10
 RyanDaileda
 Sets

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