Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Modern Algebra
January 26, 2017
Homework (Functions)
Let A and B be some sets. A rule f which to every element x ∈ A assigns a unique element y ∈ B is called a
function from A to B.
To indicate this connection between x and y we usually write y = f(x).
Recall. If A = B = {x1 , x2 , . . . , xn } for some
then every function f : A → A can be represented as
x1
x2 · · ·
f=
f(x1 ) f(x2 ) · · ·
n ∈ N, (i.e.
follows
xn
f(xn )
Example. There are exactly four functions f : A → A where A = {1, 2}
1 2
1 2
1 2
f1 =
, f2 =
, f3 =
,
1 1
1 2
2 1
f4 =
1
2
A is a finite set),
2
2
Here f3 (1) = 2, f3 (2) = 1.
I. Determine all functions f : A → A where A = {1, 2, 3}, and write them in the form
1
2
3
f=
f(1) f(2) f(3)
Example.
The function
g : {1, 2, 3} → {1, 2, 3} with g(1) = 2, g(2) = 1, g(3) = 2 can be written as follows
1 2 3
g=
.
2 1 2
II. Let ∗ be the operation on G = {e, b, c, d, f, g} defined by the operation table
∗
e
b
c
d
f
g
e
e
b
c
d
f
g
b
b
c
d
f
g
e
c
c
d
f
g
e
b
d
d
f
g
e
b
c
f
f
g
e
b
c
d
hG, ∗i is an abelian group with the identity element e
g
g
e
b
c
d
f
(based on the homework 01/17/2016).
(1) Let ξ ∈ G, and fξ : G → G is defined by fξ (x) =
ξ∗x. Determine the functions fξ : G → G
e
b
c
d
f
g
for all ξ ∈ G, and write them in the form fξ =
fξ (e) fξ (b) fξ (c) fξ (d) fξ (f ) fξ (g)
Example. For ξ = c we have
e
b
c
d
f
fc =
fc (e) fc (b) fc (c) fc (d) fc (f )
g
fc (g)
=
e
c∗e
b
c∗b
c
c∗c
d
c∗d
f
c∗f
g
c∗g
=
e
c
b
d
c
f
d
g
f
e
g
b
(2) Let fn : G → Gbe given by fn (x) = xn . Write the functions
fn for n = 1, 2, 3, 4, 5, 6 in
e
b
c
d
f
g
the form fn =
fn (e) fn (b) fn (c) fn (d) fn (f ) fn (g)
Example.
For n = 3 we have
e
b
c
d
f3 =
f3 (e) f3 (b) f3 (c) f3 (d)
f
f3 (f )
g
f3 (g)
=
e
e3
b
b3
c
c3
d
d3
f
f3
g
g3
=
e
e
b
d
c
e
d
d
f
e
(3) Write the function f : G → G given by f (x) = x−1 in the form given above.
Definition. A function f : A → B is called surjective is for each y ∈ B there exists an
x ∈ A such that f (x) = y.
(last) Example. The function f3 : |{z}
G → |{z}
G given by f3 =
A
e
e
b
d
c
e
d
d
f
e
g
d
is not surjective, because
B
for c ∈ |{z}
G there is no x ∈ |{z}
G such that f3 (x) = c.
B
A
III. Which of the functions given in I., and II. are surjective?
Definition. Let f : A → B and g : B → C be some functions. The composite function
denoted by g ◦ f is a function from A to C (i.e. g ◦ f : A → C) defined as follows:
g ◦ f (x) = g(f(x)).
e b
Example. For the functions f : |{z}
G → |{z}
G and g : |{z}
G → |{z}
G given by f =
e d
A
B
B
C
e b c d f g
g=
we have g ◦ f : |{z}
G → |{z}
G and f ◦ g : |{z}
G → |{z}
G given by
b b d d f f
A
g◦f
e
(g ◦ f)(e)
e
g(f(e))
e
g(e)
=
=
=
f◦g
e
(f ◦ g)(e)
e
f(g(e))
e
f(b)
=
=
=
b
(g ◦ f)(b)
b
g(f(b))
b
g(d)
C
c
(g ◦ f)(c)
c
g(f(c))
c
g(f )
b
f(b)
c
f(d)
f
g(e)
c
(f ◦ g)(c)
c
f(g(c))
d
f(d)
=
d
(f ◦ g)(d)
d
f(g(d))
f
f(f )
f
(g ◦ f)(f )
f
g(f(f ))
g
g(g)
g
f(f )
=
g
g(f(g))
e
b
c
f
e
d
g
f(g(g))
b
d
Here (g ◦ f)(c) = g(f(c)) = g(f ) = f as well as (f ◦ g)(c) = f(g(c)) = f(d) = d.
2
b
d
f
(f ◦ g)(f )
f
f(g(f ))
d
d
A
d
(g ◦ f)(d)
d
g(f(d))
d
g(d)
b
(f ◦ g)(b)
b
f(g(b))
C
c
f
c
d
g
(g ◦ f)(g)
d
d
f
b
g
f
g
(f ◦ g)(g)
d
d
f
e
g
e
f
e
g
g
and
g
d
IV. Consider the functions from {1, 2, 3} to {1, 2, 3} given by
f0 =
1
1
2
2
3
3
, f1 =
1
2
2
1
3
3
, f2 =
1
1
2
3
3
2
, f3 =
1
3
2
1
3
2
, f4 =
1
3
2
2
3
1
, f5 =
1
2
2
3
3
1
Compute the ”composite function” table: In the position of the intersection of the row of fi
and the column of fj is the function fi ◦ fj .
Example. In the position of f1 ◦ f2 is placed f5 , because
f1 ◦ f2 =
1
f1 (f2 (1))
2
f1 (f2 (2))
◦
f0
f1
f2
f3
f4
f5
3
f1 (f2 (3))
f0
f1
=
1
f1 (1)
2
f1 (3)
f2
f3
f4
f5
3
3
f1 (2)
f5
=
1
2
2
3
3
1
= f5