Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Modern Algebra
January 31, 2017
Homework (Permutations)
Definition. Let n ∈ N, and f ∈ Sn , given by
1
2 ···
n
1 2 ···
f=
=
f(1) f(2) · · · f(n)
a1 a2 · · ·
n
an
in other words f : {1, 2, . . . , n} → {1, 2, . . . , n} is a bijective function with f(i) = ai for
each i ∈ {1, 2, . . . , n}, and {a1 , a2 , . . . , an } = {1, 2, . . . , n}.
Example. For n = 4 let f ∈ S4 be given by
1
2
3
f=
a1 a2 a3
4
a4
=
1
3
2
1
3
4
4
2
in other words f : {1, 2, 3, 4} → {1, 2, 3, 4} is a bijective function with f(1) = a1 = 3, f(2) = a2 = 1,
f(3) = a3 = 4, f(4) = a4 = 2, and {|{z}
3 , |{z}
1 , |{z}
4 , |{z}
2 } = {1, 2, 3, 4}.
a1
a2
a3
a4
The function f−1 : {1, 2, . . . , n} → {1, 2, . . . , n} given by f−1 (ai ) = i is called the inverse
function of f.
Example. The inverse function of

1
2
3
1
f =  |{z}
|{z}
a1
a2

4
2 
|{z}
3
4
|{z}
a3
is
f−1 =
1
2
2
4
3
1
4
3
a4
because f−1 ( a1 ) = 1, f−1 ( a2 ) = 2, f−1 ( a3 ) = 3, and f−1 ( a4 ) = 4.
|{z}
|{z}
|{z}
|{z}
3
1
4
2
···
n
:
· · · an
a1 a2 · · · an
Step 1. Interchange the rows representing f, i.e.
1
2 ···
n
a1 a2 · · · an
Step 2. Interchange the columns of
such that the integers in the top row are increasing
1
2 ···
n
−1
from 1 to n. The result represents the function f .


1
2
3
4
3
1
4
2 
Example. The construction of the inverse function f−1 for f =  |{z}
|{z}
|{z}
|{z}
A way to construct the inverse of f =
1
a1
2
a2
a1
1
3
2
1
3
4
4
2
Step 1
3
1
1
2
4
3
2
4
a2
Step 2
a3
1
2
2
4
2
1
3
2
a4
3
1
4
3
1
3
= f−1
I. Consider the permutations from S3 :
f0 =
1
1
2
2
3
3
, f1 =
1
2
2
1
3
3
, f2 =
1
1
2
3
3
2
, f3 =
1
3
, f4 =
2
2
3
1
, f5 =
1
2
2
3
3
1
Determine the inverse of each function fi :
Remark. For each f ∈ Sn we have f◦f−1 =
Check that fi ◦ f−1
=
i
1
1
2
2
3
3
1
1
2
2
and f−1
◦ fi =
i
···
···
i
i
1
1
2
2
f−1
5 =
f−1
4 =
f−1
3 =
f−1
2 =
f−1
1 =
f−1
0 =
··· n
1
and f−1 ◦f =
··· n
1
3
for all i ∈ {0, . . . , 5}
3
2
2
···
···
i
i
···
···
n
n
Notation. Let n ∈ N, and f ∈ Sn . For any positive integer m we will use the following
notations
(1) fm = |f ◦ f ◦{z· · · ◦ }f,
m−times
−m
(2) f
−1
= f|
◦ f−1{z
◦ · · · ◦ f−1},
m−times
n
n
2 3
1 2 3
1 2 3
1 2 3
1
◦
◦
◦
=
1 3
3 2 1
3 2 1
2 3 1
1
{z
} |
{z
} |
{z
} |
{z
}
2
3
0
0
(3) For any f ∈ Sn the function f is equal to f =
1 2 ···
1 2 ···
i ···
i ···
II. Let f0 , f1 , f2 , f3 , f4 , f5 be from the last exercise. Compute the following:
(1) f1 ◦ f−2
4 ◦ f5
−1
−1
(2) f3 ◦ f4 ◦ f5 ◦ f−1
5 ◦ f4 ◦ f3
−1
−1
(3) f3 ◦ f4 ◦ f5 ◦ f−1
3 ◦ f4 ◦ f5
36
(4) f−1
0
(5) f05
(6) f1 ◦ f−2
4 ◦ f5
−1
3
(7) f21 ◦ f−2
4 ◦ f5
−3
Example. f1 ◦
f−2
4
◦ f5 =
|
1
2
f−1
4 =f4
f1
f−1
4 =f4
1 2 3 4 5 6
2 3 4 1 6 5
.
3
2
= f2
f5
II. Let f ∈ S6 be given by f =
n
o n
o
Show that the set H = fn : n ∈ {1, 2, . . . , 317} = f, f2 , f3 , . . . , f317 has four elements.
2