Homework 2 Solutions
1. Suppose
α
∈
S
n
is an
r
cycle. Let
α
= (
x
1
x
2
. . . x
r
) where
x
i
6
=
x
j
for
i
6
=
j
.
(a)
α
only permutes the elements
x
1
through
x
r
and leaves any other elements from
{
1
, . . . n
}
ﬁxed. Thus in order to show that
α
r
= (1) it suﬃces to show that
α
r
(
x
i
) =
x
i
for each
1
≤
i
≤
r
. To see this observe:
x
2
=
α
(
x
1
)
x
3
=
α
(
x
2
) =
α
2
(
x
1
)
.
.
.
x
r
=
α
(
x
r

1
) =
α
r

1
(
x
1
)
x
1
=
α
(
x
r
) =
α
r
(
x
1
).
Similarly for all
i
, with 1
≤
i
≤
r
,
x
i
=
α
r
(
x
i
). Thus
α
r
= (1).
(b) BWOC suppose
∃
k
such that 0
< k < r
and
α
k
= (1). By part (a) we know
x
k
+1
=
α
k
(
x
1
). Since
α
k
= (1) by our supposition, this implies that
x
k
+1
=
x
1
. Since
k < r
,
1
< k
+ 1
≤
r
. Thus
x
1
=
x
k
+1
for 1
< k
+ 1
≤
r
. CONTRADICTION
x
i
’s are distinct
for 1
≤
i
≤
r
.
2. Let
α
= (
x
1
x
2
. . . x
r
) be an
r
cycle.
(
⇒
) Suppose
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 Spring '11
 Johnson
 Algebra, Group Theory, Permutations, Parity, GFI, x1 x2

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