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Math 113 Homework # 4, selected solutions
2:
(a) Let
i
∈ {
1
,...,n
}
; we must show that
f
(
g
(
i
)) =
g
(
f
(
i
)).
Case 1:
i
is not in the support of
f
or
g
. Then
f
(
g
(
i
)) =
f
(
i
) =
i
and likewise
g
(
f
(
i
)) =
g
(
i
) =
i
.
Case 2:
i
is in the support of
f
. Then
f
(
i
) is also in the support
of
f
, since otherwise
f
would send both
i
and
f
(
i
) to
f
(
i
), and
hence would not be injective since
i
6
=
f
(
i
). Since
i
is not in the
support of
g
we have
f
(
g
(
i
)) =
f
(
i
), and since
f
(
i
) is not in the
support of
g
we have
g
(
f
(
i
)) =
f
(
i
).
Case 3:
i
is in the support of
g
. By the same argument as in Case
2 but with
f
and
g
switched, we have
f
(
g
(
i
)) =
g
(
f
(
i
)) =
g
(
i
).
(b) Let
p
be the element of supp(
f
)
∩
supp(
g
). Then you can check
that [
f,g
](
p
) =
f
(
p
), [
f,g
](
f
(
p
)) =
g
(
p
), [
f,g
](
g
(
p
)) =
p
, and for
any
x
∈ {
1
,...,n
}\{
p,f
(
p
)
,g
(
p
)
}
, we have [
f,g
](
x
) =
x
. Also
p
,
f
(
p
), and
g
(
p
) are all distinct (why?), so [
f,g
] is the 3cycle
[
f,g
] = (
p, g
(
p
)
, f
(
p
))
.
3:
(a) Let
i
∈ {
1
,...,n
}
; we need to show that
σμσ

1
(
i
) = (
σ
(
x
1
)
σ
(
x
2
)
···
σ
(
x
k
))(
i
)
.
We consider two cases.
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 Fall '08
 OGUS
 Math

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