Worked Examples on Groups
Here are some worked examples in group theory.
Exercise 1
Does the finite set A = { -3, -2, -1, 0, 1, 2, 3 } form a group under integer addition (+)?
To answer this question, we will examine whether the set A, equipped with the operation of addition, satisfies all the group axioms.
- Associativity
Addition on A is associative: $$ (a + b) + c = a + (b + c) \ \ \ \ \forall \ a, b, c \in A $$ - Identity element
The set A contains an additive identity: 0. For any element a in A: $$ a + 0 = 0 + a = a $$ - Inverse element
Every element in A has an additive inverse within A. For example: $$ 1 + (-1) = 0, \quad 2 + (-2) = 0, \quad 3 + (-3) = 0, \quad 0 + 0 = 0 $$ - Closure
The set A is not closed under integer addition. In order for A to form a group under addition, the sum of any two elements in A must also belong to A. This condition fails. For example: $$ 3 + 1 = 4 \notin A $$
Although the first three axioms are satisfied, the problem explicitly asks whether A forms a group under ordinary integer addition.
Note. This is ordinary integer addition, not modular arithmetic.
Conclusion
Therefore, the set A does not form a group under integer addition.
Exercise 2
Does the set of integers Z form a group under multiplication?
Let us examine whether the group axioms are satisfied.
- Closure
The set Z is closed under multiplication: the product of any two integers is again an integer. $$ \forall a, b \in Z \ \Rightarrow \ a \cdot b \in Z $$ - Associativity
Multiplication on Z is associative: $$ (a \cdot b) \cdot c = a \cdot (b \cdot c) \ \ \ \ \forall \ a, b, c \in Z $$ - Identity element
The integer 1 serves as the multiplicative identity in Z: $$ a \cdot 1 = 1 \cdot a = a $$ - Inverse element
Not every integer has a multiplicative inverse in Z. For instance, the multiplicative inverse of 3 is 1/3, which is not an integer: $$ 3 \cdot \frac{1}{3} = 1 $$ Thus, the inverse axiom is not satisfied.
Conclusion
In conclusion, the set of integers Z does not form a group under multiplication.
Exercise 3
Is the set of integers Z a group under addition?
To answer this question, we verify whether the group axioms are satisfied.
- Closure
The set Z is closed under addition: the sum of any two integers is an integer. $$ \forall a, b \in Z \ \Rightarrow \ a + b \in Z $$ - Associativity
Addition on Z is associative: $$ (a + b) + c = a + (b + c) \ \ \ \ \forall \ a, b, c \in Z $$ - Identity element
Z contains an additive identity, namely 0: $$ a + 0 = 0 + a = a $$ for all a in Z. - Inverse element
Every integer a has an additive inverse −a in Z: $$ a + (-a) = (-a) + a = 0 $$
All group axioms are satisfied.
Conclusion
Therefore, the set of integers Z, equipped with addition, forms a group (Z, +).
Exercise 4
Does the set A = { 1, −1, i, −i }, consisting of four complex numbers, form a group under multiplication?
We proceed to verify whether the group axioms are satisfied.
- Closure
Multiplication is closed on the set A = { 1, −1, i, −i }: $$ \forall a, b \in A \ \Rightarrow \ a \cdot b \in A $$Verification. Recall that in the complex numbers, the imaginary unit satisfies i2 = −1.
$$ 1 \cdot 1 = 1 \in A $$ $$ 1 \cdot (-1) = -1 \in A $$ $$ -1 \cdot (-1) = 1 \in A $$ $$ i \cdot i = i^2 = -1 \in A $$ $$ i \cdot (-i) = -i^2 = -(-1) = 1 \in A $$ $$ -i \cdot (-i) = i^2 = -1 \in A $$ $$ 1 \cdot i = i \in A $$ $$ -1 \cdot i = -i \in A $$ - Associativity
Multiplication on A is associative: $$ (a \cdot b) \cdot c = a \cdot (b \cdot c) \ \ \ \ \forall \ a, b, c \in A $$ - Identity element
A contains a multiplicative identity, namely 1: $$ a \cdot 1 = 1 \cdot a = a \ \ \ \forall \ a \in A $$ - Inverse element
Every element of A = { 1, −1, i, −i } has a multiplicative inverse in A: $$ a \cdot a^{-1} = 1 \ \ \ \forall \ a \in A $$ Here, a−1 is the multiplicative inverse within the set A - that is, an element of A such that a · a−1 = 1.
Verification.
The inverse of 1 is 1: $$ 1 \cdot 1 = 1 $$
The inverse of −1 is −1: $$ -1 \cdot -1 = 1 $$
The inverse of i is −i: $$ i \cdot (-i) = -i^2 = -(-1) = 1 $$
The inverse of −i is i: $$ -i \cdot i = -i^2 = -(-1) = 1 $$
All group axioms are satisfied.
Conclusion
Therefore, the set A, under multiplication, forms a group (A, ⋅).
Exercise 5
Verify whether the finite set of integers Z8 = { 0, 1, 2, 3, 4, 5, 6, 7 } forms a group under addition modulo 8 (+8).
- Closure
Addition modulo 8 (+8) is closed on Z8: the sum of any two elements in Z8, computed modulo 8, remains in Z8. The Cayley table for (Z8, +8) is as follows:
a +8 b 0 1 2 3 4 5 6 7 0 0 1 2 3 4 5 6 7 1 1 2 3 4 5 6 7 0 2 2 3 4 5 6 7 0 1 3 3 4 5 6 7 0 1 2 4 4 5 6 7 0 1 2 3 5 5 6 7 0 1 2 3 4 6 6 7 0 1 2 3 4 5 7 7 0 1 2 3 4 5 6 - Associativity
Addition modulo 8 is associative on Z8: $$ (a + b) + c = a + (b + c) \ \ \ \ \forall \ a, b, c \in Z_8 $$ - Identity element
Z8 contains an identity element for addition modulo 8, namely 0: $$ a + 0 = 0 + a = a \ \ \ \forall \ a \in Z_8 $$ - Inverse element
Every element a in Z8 has an additive inverse a−1 in Z8, such that: $$ a + a^{-1} = a^{-1} + a = 0 $$Example. The additive inverse of a = 7 is a−1 = 1, since: $$ 7 +_8 1 = 0 $$
The additive inverse of a = 6 is a−1 = 2, since: $$ 6 +_8 2 = 0 $$
And so on.
All group axioms are satisfied.
Conclusion
Therefore, the finite set Z8 forms a group under addition modulo 8.
Exercise 6
Determine the proper and improper subgroups of the multiplicative group (S, ·), where S = { 1, −1, i, −i } consists of four complex numbers.
The multiplicative group (S, ·) is defined on the following set of complex numbers:
$$ S = \{ 1, -1, i, -i \} $$
Here, i denotes the imaginary unit in the field of complex numbers.
We begin by listing all subsets of S - that is, its power set.
As a first step, we eliminate all subsets S' ⊆ S that do not contain the identity element u = 1, since such subsets cannot form subgroups of (S, ·).
1 | -1 | i | -i | S' ⊆ S | Nota |
---|---|---|---|---|---|
* | * | * | * | S'={1,-1,i,-i} | |
* | * | * | S'={1,-1,i} | ||
* | * | * | S'={1,-1,-i} | ||
* | * |
|
S'={1,-1} | ||
* | * | * | S'={1,i,-i} | ||
* | * | S'={1,i} | |||
* | * | S'={1,-i} | |||
* | S'={1} | ||||
* | * | * | S'={-1,i,-i} | does not contain the identity element u = 1 | |
* | * | S'={-1,i} | does not contain the identity element u = 1 | ||
* | * | S'={-1,-i} | does not contain the identity element u = 1 | ||
* | S'={-1} | does not contain the identity element u = 1 | |||
* | * | S'={i,-i} | does not contain the identity element u = 1 | ||
* | S'={i} | does not contain the identity element u = 1 | |||
* | S'={-i} | does not contain the identity element u = 1 | |||
S'={} | does not contain the identity element u = 1 |
We then identify the improper subgroups: the entire group S' = S, and the trivial subgroup S' = { 1 }:
1 | -1 | i | -i | S' ⊆ S | Nota |
---|---|---|---|---|---|
* | * | * | * | S'={1,-1,i,-i} | improper subgroups S'=S |
* | * | * | S'={1,-1,i} | ||
* | * | * | S'={1,-1,-i} | ||
* | * |
|
S'={1,-1} | ||
* | * | * | S'={1,i,-i} | ||
* | * | S'={1,i} | |||
* | * | S'={1,-i} | |||
* | S'={1} | trivial subgroup S'={u}={1} |
Next, we examine the remaining candidate subsets S' to check whether they are closed under multiplication. Since: $$ i^2 = -1 $$ we verify closure for each subset:
1 | -1 | i | -i | S' ⊆ S | Nota |
---|---|---|---|---|---|
* | * | * | * | S'={1,-1,i,-i} | improper subgroups S'=S |
* | * | * | S'={1,-1, i} | -1·i = -i ∉ {1,-1,i} | |
* | * | * | S'={1,-1,-i} | -1·(-i) = i ∉ {1,-1,-i} | |
* | * |
|
S'={1,-1} | ||
* | * | * | S'={1,i,-i} | i·i = i2=-1 ∉ {1,i,.i} | |
* | * | S'={1,i} | i·i = i2=-1 ∉ {1,i} | ||
* | * | S'={1,-i} | (-i)·(-i) = i2=-1 ∉ {1,-i} | ||
* | S'={1} | trivial subgroup S'={u}={1} |
We now focus on the subset S' = { 1, −1 } to determine whether it forms a subgroup:
- S' = { 1, −1 } contains the identity element u = 1.
- S' = { 1, −1 } is closed under multiplication.
It remains to verify the remaining group axioms:
- Every element in S' has an inverse within S': $$ 1 \cdot 1 = 1 = u $$ $$ (-1) \cdot (-1) = 1 = u $$
- Multiplication is associative on S': $$ (a \cdot b) \cdot c = a \cdot (b \cdot c) \ \ \ \ \forall \ a, b, c \in \{ 1, -1 \} $$
All group axioms are satisfied.
We conclude that the subset S' = { 1, −1 } is a proper subgroup of (S, ·).
1 | −1 | i | −i | S' ⊆ S | Note |
---|---|---|---|---|---|
* | * | * | * | S' = { 1, −1, i, −i } | improper subgroup S' = S of (S, ·) |
* | * | S' = { 1, −1 } | proper subgroup of (S, ·) | ||
* | S' = { 1 } | trivial subgroup S' = { u } = { 1 } of (S, ·) |
In conclusion, the multiplicative group (S, ·) has precisely three subgroups, two of which are improper.
Exercise 6bis
Determine the generators of the cyclic group (Z8, +8), where Z8 = { 0, 1, 2, 3, 4, 5, 6, 7 }, under addition modulo 8.
We examine the cyclic subgroups generated by each element of Z8:
Element x | Subgroup generated by x | Order of x |
---|---|---|
<0> | { 0 } | 1 |
<1> | { 0, 1, 2, 3, 4, 5, 6, 7 } | 8 |
<2> | { 0, 2, 4, 6 } | 4 |
<3> | { 3, 6, 1, 4, 7, 2, 5, 0 } | 8 |
<4> | { 0, 4 } | 2 |
<5> | { 5, 2, 7, 4, 1, 6, 3, 0 } | 8 |
<6> | { 6, 4, 2, 0 } | 4 |
<7> | { 7, 6, 5, 4, 3, 2, 1, 0 } | 8 |
The elements 1, 3, 5, and 7 each generate the entire group Z8. Therefore, these elements are generators of the cyclic group (Z8, +8).
- <1> = { 0, 1, 2, 3, 4, 5, 6, 7 } = Z8
- <3> = { 3, 6, 1, 4, 7, 2, 5, 0 } = Z8
- <5> = { 5, 2, 7, 4, 1, 6, 3, 0 } = Z8
- <7> = { 7, 6, 5, 4, 3, 2, 1, 0 } = Z8
Exercise 7
Determine whether the symmetric group on the set S = { 1, 2, 3, 4 } is an abelian group.
The set S = { 1, 2, 3, 4 } consists of n = 4 elements:
$$ S = \{ 1, 2, 3, 4 \} $$
The symmetric group Sn is the group of all permutations of S, with composition of permutations as the group operation.
The set S4 contains n! = 24 elements.
σ | Permutazione | Notazione alternativa |
---|---|---|
σ1 | 1,2,3,4 | (1) |
σ2 | 2,1,3,4 | (1,2) |
σ3 | 3,2,1,4 | (1,3) |
σ4 | 4,2,3,1 | (1,4) |
σ5 | 1,3,2,4 | (2,3) |
σ6 | 1,4,3,2 | (2,4) |
σ7 | 1,2,4,3 | (3,4) |
σ8 | 2,3,1,4 | (1,2,3) |
σ9 | 3,1,2,4 | (1,3,2) |
σ10 | 2,4,3,1 | (1,2,4) |
σ11 | 4,1,3,2 | (1,4,2) |
σ12 | 3,2,4,1 | (1,3,4) |
σ13 | 4,2,1,3 | (1,4,3) |
σ14 | 1,3,4,2 | (2,3,4) |
σ15 | 1,4,2,3 | (2,4,3) |
σ16 | 2,3,4,1 | (1,2,3,4) |
σ17 | 2,4,1,3 | (1,2,4,3) |
σ18 | 4,1,2,3 | (1,4,3,2) |
σ19 | 3,1,4,2 | (1,3,4,2) |
σ20 | 3,4,2,1 | (1,3,2,4) |
σ21 | 4,3,1,2 | (1,4,2,3) |
σ22 | 3,4,1,2 | (1,3)(2,4) |
σ23 | 4,3,2,1 | (1,4)(2,3) |
σ24 | 2,1,4,3 | (1,2)(3,4) |
Example 1. The permutation σ8 = (2, 3, 1, 4) can also be expressed in two-line notation as: $$ \sigma_8 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 1 & 4 \end{pmatrix} $$ In functional terms: $$ \sigma_8 : S \rightarrow S \\ 1 \mapsto 2 \\ 2 \mapsto 3 \\ 3 \mapsto 1 \\ 4 \mapsto 4 $$ Since the permutation consists of a 3-cycle (1 → 2 → 3 → 1) and the identity on 4, it can also be written in cycle notation as: $$ (1 \ 2 \ 3) $$ The identity cycle on 4 is omitted, as is customary.
Example 2. The permutation σ22 = (3, 4, 1, 2) is written as: $$ \sigma_{22} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix} $$ That is: $$ \sigma_{22} : S \rightarrow S \\ 1 \mapsto 3 \\ 2 \mapsto 4 \\ 3 \mapsto 1 \\ 4 \mapsto 2 $$ which corresponds to two disjoint 2-cycles: $$ (1 \ 3)(2 \ 4) $$
The set of permutations S4 forms a symmetric group, denoted (S4, o), under the operation of composition o.
$$ (S_4,\text{o}) $$
Here, the composition operation refers to the composition of functions, written as fog = f(g), where f and g are permutations.
For example, the composition σ3 o σ5 = σ3(σ5) is given by:
$$ \sigma_3 \ \text{o} \ \sigma_5 = (3,2,1,4) \ \text{o} \ (1,3,2,4) $$
which can also be expressed as:
$$ \sigma_3 \ \text{o} \ \sigma_5 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4 \end{pmatrix} \ \text{o} \ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix} $$
Note. In the alternative cycle notation for permutations, this composition is written as: $$ \sigma_3 \text{o} \sigma_5 = (1,3) \ \text{o} \ (2,3) $$
In the composition fog = f(g), we take g = σ5 = (1,3,2,4) and f = σ3 = (3,2,1,4).
We apply the inner permutation g first, followed by the outer permutation f:
$$ 1 \rightarrow_g 1 \rightarrow_f 3 $$
$$ 2 \rightarrow_g 3 \rightarrow_f 1 $$
$$ 3 \rightarrow_g 2 \rightarrow_f 2 $$
$$ 4 \rightarrow_g 4 \rightarrow_f 4 $$
The result of this composition is the permutation (3,1,2,4):
$$ \sigma_3 \ \text{o} \ \sigma_5 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4 \end{pmatrix} \ \text{o} \ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 2 & 4 \end{pmatrix} $$
That is:
$$ \sigma_3 \ \text{o} \ \sigma_5 = (3,2,1,4) \ \text{o} \ (1,3,2,4) = (3,1,2,4) $$
Note. In the alternative cycle notation, the result is written as: $$ \sigma_3 \ \text{o} \ \sigma_5 = (1,3) \ \text{o} \ (1,3,2) $$
The resulting permutation σ3 o σ5 is (3,1,2,4), which is an element of Sn:
$$ (3,1,2,4) = \sigma_9 \in S_n $$
To determine whether (Sn, o) forms an abelian group, we need to verify whether the composition is commutative.
Let’s compute the reverse composition σ5 o σ3 = σ5(σ3):
$$ \sigma_5 \ \text{o} \ \sigma_3 = (1,3,2,4) \ \text{o} \ (3,2,1,4) $$
which can be written as:
$$ \sigma_5 \ \text{o} \ \sigma_3 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix} \ \text{o} \ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4 \end{pmatrix} $$
Again, we apply the inner permutation g = σ3 first, followed by f = σ5:
$$ 1 \rightarrow_g 4 \rightarrow_f 2 $$
$$ 2 \rightarrow_g 2 \rightarrow_f 3 $$
$$ 3 \rightarrow_g 1 \rightarrow_f 1 $$
$$ 4 \rightarrow_g 4 \rightarrow_f 4 $$
The resulting permutation is (2,3,1,4):
$$ \sigma_5 \ \text{o} \ \sigma_3 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix} \ \text{o} \ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 1 & 4 \end{pmatrix} $$
That is:
$$ \sigma_5 \ \text{o} \ \sigma_3 = (1,3,2,4) \ \text{o} \ (3,2,1,4) = (2,3,1,4) $$
The result of this composition σ5 o σ3 is (2,3,1,4), which is also an element of Sn:
$$ (2,3,1,4) = \sigma_8 \in S_n $$
Conclusion
The composition σ3 o σ5 yields (3,1,2,4):
$$ \sigma_3 \ \text{o} \ \sigma_5 = (3,2,1,4) \ \text{o} \ (1,3,2,4) = (3,1,2,4) $$
Whereas the reverse composition σ5 o σ3 yields (2,3,1,4):
$$ \sigma_5 \ \text{o} \ \sigma_3 = (1,3,2,4) \ \text{o} \ (3,2,1,4) = (2,3,1,4) $$
Since the two results differ, the operation is not commutative:
$$ \sigma_3 \ \text{o} \ \sigma_5 \ne \sigma_5 \ \text{o} \ \sigma_3 $$
In conclusion, the group (S4, o) is non-abelian because it fails to satisfy the commutative property.
Note. In fact, all symmetric groups Sn with more than two elements (n > 2) are non-abelian - that is, they are non-commutative groups.
Exercise 8
Determine whether the additive group (R,+) is homomorphic to itself under the mapping f: x→ x2
$$ f:(R,+) \rightarrow (R,+) $$
A group homomorphism from a group (G, *) to a group (H, #) is a function f such that:
$$ f:G \rightarrow H $$
and for all elements a, b in G, the following property holds:
$$ \forall \ a,b \in G \Rightarrow f(a * b) = f(a) \# f(b) $$
In this case, we are considering the function f(x) = x2, with G = H = R, and both groups being (R, +).
Thus, the binary operations involved are * = + and # = +.
$$ \forall \ a,b \in R \Rightarrow (a+b)^2 = a^2 + b^2 $$
However, the square of a sum is not equal to the sum of the squares:
$$ (a+b)^2 \ne a^2 + b^2 $$
since:
$$ (a+b)^2 = a^2 + b^2 + 2ab $$
Therefore, the additive group (R, +) is not homomorphic to itself under the mapping f(x) = x2.
Exercise 9
Determine whether the additive group (Z, +) is homomorphic to the multiplicative group (Z, ·) under the mapping f: x→ ix
where i is the imaginary unit in the complex numbers.
$$ f:(Z,+) \rightarrow (Z, \cdot) $$
A group homomorphism from a group (G, *) to a group (H, #) is a function f such that:
$$ f:G \rightarrow H $$
and for all elements a, b in G, the following condition is satisfied:
$$ \forall \ a,b \in G \Rightarrow f(a * b) = f(a) \# f(b) $$
Here, the function is f(x) = ix, with G = Z and H = Z, and the groups being (Z, +) and (Z, ·), respectively.
Thus, the binary operations are * = + and # = ·.
$$ \forall \ a,b \in Z \Rightarrow i^{a+b} = i^a \cdot i^b $$
By the laws of exponents:
$$ i^{a+b} = i^a \cdot i^b $$
We see that the condition f(a * b) = f(a) # f(b) holds for all integers a, b ∈ Z.
Therefore, the additive group (Z, +) is homomorphic to the multiplicative group (Z, ·) under the map f(x) = ix.
Example. Consider the integers a = 2 and b = 3 in the additive group (Z, +). Let us verify:
$$ f(a + b) = f(a) \cdot f(b) $$
The mapping is f: x → ix.
$$ i^{(2+3)} = i^2 \cdot i^3 $$
$$ i^5 = i^5 $$
The equality holds - and it holds for any integers a and b.
And so on.