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Introduction to Group Theory
Solutions 2
Unit website: http://www.maths.bris.ac.uk/~majcr/IGT.html
Questions 1,2,3 to be handed in for tutorials.
1. Which of the following are groups, and why? For those that are
groups, find the identity element, find the inverses of all the elements,
and decide whether the group is abelian.
(i) The set of even integers under addition.
(ii) The set of odd integers under addition.
(iii) The set {x ∈ R : x > 0} of positive real numbers under multiplication.
(iv) (Z, ?), where x ? y = x + y + 1.
(v) (Z, •), where x • y = y for all x, y ∈ Z.
Solution:
(i) It is a group. If 2a and 2b are even integers, then 2a + 2b =
2(a + b) is even, so the set is closed under addition. Addition
is associative. 0 = 2.0 is an even integer, and is an identity
for addition. If 2a is an even integer, then so is −2a, which is
an inverse to 2a. It is abelian, since addition is commutative
(2a + 2b = 2b + 2a).
(ii) Not a group. The odd integers are not closed under addition
(for example, 1 + 1 = 2, which is even). Note that to show
something is not a group then it is enough to find one of the
necessary properties that fails.
(iii) It is a group. The product of two positive real numbers is also
positive, so the set is closed under multiplication. Multiplication is associative. 1 is a positive real number, and is an
identity for multiplication. If x is a positive real number, so is
1
, which is an inverse for x. It is abelian, since multiplication
x
is commutative.
(iv) It is a group. If x and y are integers, so is x + y + 1, so Z is
closed under the operation ?. If x, y, z ∈ Z, then
(x ? y) ? z = (x ? y) + z + 1 = (x + y + 1) + z + 1 = x + y + z + 2,
and
x ? (y ? z) = x + (y ? z) + 1 = x + (y + z + 1) + 1 = x + y + z + 2
is the same, so the operation ? is associative. −1 is an identity,
since
x ? −1 = x + (−1) + 1 = x = (−1) + x + 1 = −1 ? x
for any x ∈ Z. The inverse of x is −x − 2, since
x ? (−x − 2) = x + (−x − 2) + 1 = −1
is the identity, as is
(−x − 2) ? x = (−x − 2) + x + 1 = −1.
It is abelian, since
x?y =x+y+1=y+x+1=y?x
for all x, y ∈ Z.
(v) It is not a group, since it does not have an identity: since x•e =
e for all x, e ∈ Z, there is no element e for which x • e = x for
all x ∈ Z (it is true that, for any e, e • x = x, but for an identity
element we also need x • e = x).
2. Which of the following sets of functions f : R → R are groups under
composition, and why?
(i) The set of all functions R → R.
(ii) The set of all functions of the form f (x) = ax + b, where a and
b are real numbers.
(iii) The set of all functions of the form f (x) = ax + b, where a and
b are real numbers and a 6= 0.
Solution:
(i) Not a group, since not every function has an inverse (e.g., the
zero function given by f (x) = 0 for all x ∈ R does not have an
inverse).
(ii) Not a group, since (taking a = b = 0) the zero function is of
this form and does not have an inverse.
(iii) It is a group. Let fa,b (x) = ax + b when a, b ∈ R. Then
fa,b (fc,d (x)) = a(cx + d) + b = acx + (ad + b) = fac,ad+b (x)
and if a 6= 0 and c 6= 0 then ac 6= 0, so the set is closed under
composition. Composition of functions is associative. f1,0 is the
identity function. The inverse of fa,b is f 1 ,− b , since
a
a
x b
fa,b f 1 ,− b (x) = a( − ) + b = x
a
a
a a
and
1
b
f 1 ,− b (fa,b (x)) = (ax + b) − = x.
a
a
a
a
3. Let G = {e, x, y} be a set with three elements. Show that there is
exactly one binary operation making G a group with identity element
e.
Solution: We’ll write the group multiplicatively. We must have
ee = e, xe = x, ye = y, ex = x and ey = y by the property of an
identity.
Consider xy. If xy = x then y = x−1 xy = x−1 x = e giving a
contradiction. If xy = y then x = xyy −1 = yy −1 = e, again giving a
contradiction. So the only possibility is xy = e. Similarly yx = e.
Now consider x2 . If x2 = x then x = x2 x−1 = xx−1 = e giving a
contradiction. If x2 = e then x2 = xy and so x = x−1 x2 = x−1 xy = y,
again giving a contradiction. So the only possibility is x2 = y. Similarly
y 2 = x.
So the Cayley (multiplication) table of G is
e
e e
x x
y y
x
x
y
e
y
y
e
x
4. Let G be the symmetry group of a circle. You may assume that
every symmetry is either a rotation around the centre of the circle, or
a reflection in a line through the centre of the circle.
For every pair (R, S) of symmetries, calculate the composition R ◦ S.
Solution: Let Rθ be a rotation anticlockwise by an angle of θ, and
let Sθ be a reflection in the line obtained by rotating the y-axis anticlockwise through an angle of θ.
Then Rθ Rφ = Rθ+φ .
Sφ Rθ = Sφ− θ .
2
Rθ Sφ = Sφ+ θ .
2
Sφ Sθ = R2φ−2θ .
5. The “Sudoku group” consists of the permutations of the 81 squares
in a Sudoku grid that will transform every legally completed Sudoku
grid into another legally completed Sudoku grid. (For example, swapping the first two rows.) Show that this is a group under composition,
and try to list as many different kinds of elements as you can.
Solution: Suppose A and B are elements of the “Sudoku group”,
and we start with a legally completed Sudoku grid. If we apply B then
we get another legal grid, and so if we apply A to that then we get yet
another legal grid. So in other words AB takes legal grids to legal grids,
and so AB is in the Sudoku group, and so the group is closed under
composition. Composition of permutations is an associative operation.
Clearly the identity is in the group and the inverse of an element is
also in the group.
Here are some elements of the group:
We can permute the first three rows in any way we like. Or the
fourth to sixth rows. Or the seventh to ninth rows. Or the same with
columns.
We can permute blocks of three rows. For example, swap the block
consisting of the first three rows with the block consisting of the fourth
to sixth rows. Or the same with columns.
We can reflect in the main diagonal (so that rows become columns
and vice versa).
In total, it turns out that this group has just over three million
(3359232, to be precise) elements, all of which can be obtained by
composing the simple elements described above.
c
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