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Homework V: July 10, 2009
2
.
25
(
i
)
.
If
α
is an
r
cycle, show that
α
r
= (1).
Write
α
= (
i
0
i
1
. . . i
r

1
). The proof of Lemma 2.25(ii) shows that
α
k
(
i
0
) =
i
k
, where subscripts are read mod
r
. It follows that
α
r
(
i
0
) =
i
0
. But we know
that we can also write
α
= (
i
j
i
j
+1
. . . i
j

1
) for any
j
, so that
α
r
(
i
j
) =
i
j
.
Therefore,
α
r
ﬁxes all
i
j
. Since
α
ﬁxes any other integers between 1 and
n
, if
any, so does
α
r
. Therefore,
α
r
= (1), for it ﬁxes everything.
2
.
25
(
ii
)
.
If
α
is an
r
cycle, show that
r
is the least positive integer such that
α
r
= (1).
If 0
< k < r
, then
α
k
(
i
0
) =
i
k
6
=
i
0
, and so
α
k
6
= (1).
2
.
30
(
i
)
.
Prove that if
α, β
are (not necessarily disjoint) permutations that com
mute, then (
αβ
)
k
=
α
k
β
k
for all
k
≥
1.
We prove ﬁrst, by induction on
n
≥
1, that
βα
n
=
α
n
β
. The base step
n
= 1
is true because we are assuming that
α
and
β
commute. For the inductive step,
βα
n
+1
=
βα
n
α
=
α
n
βα
=
α
n
αβ
=
α
n
+1
β
(the second equality is the inductive hypothesis). We now prove the desired
result by induction on
k
≥
1. The base step
k
= 1 is obviously true. For the
inductive step,
(
αβ
)
k
+1
=
αβ
(
αβ
)
k
=
αβα
k
β
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This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Algebra

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