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.

  1. Associativity
    Addition on A is associative: $$ (a + b) + c = a + (b + c) \ \ \ \ \forall \ a, b, c \in A $$
  2. Identity element
    The set A contains an additive identity: 0. For any element a in A: $$ a + 0 = 0 + a = a $$
  3. Inverse element
    Every element in A has an additive inverse within A. For example: $$ 1 + (-1) = 0, \quad 2 + (-2) = 0, \quad 3 + (-3) = 0, \quad 0 + 0 = 0 $$
  4. 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.

  1. Closure
    The set Z is closed under multiplication: the product of any two integers is again an integer. $$ \forall a, b \in Z \ \Rightarrow \ a \cdot b \in Z $$
  2. Associativity
    Multiplication on Z is associative: $$ (a \cdot b) \cdot c = a \cdot (b \cdot c) \ \ \ \ \forall \ a, b, c \in Z $$
  3. Identity element
    The integer 1 serves as the multiplicative identity in Z: $$ a \cdot 1 = 1 \cdot a = a $$
  4. 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.

  1. Closure
    The set Z is closed under addition: the sum of any two integers is an integer. $$ \forall a, b \in Z \ \Rightarrow \ a + b \in Z $$
  2. Associativity
    Addition on Z is associative: $$ (a + b) + c = a + (b + c) \ \ \ \ \forall \ a, b, c \in Z $$
  3. Identity element
    Z contains an additive identity, namely 0: $$ a + 0 = 0 + a = a $$ for all a in Z.
  4. 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:

  1. S' = { 1, −1 } contains the identity element u = 1.
  2. 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 = σ35) 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 = σ53):

$$ \sigma_5 \ \text{o} \ \sigma_3 = (1,3,2,4) \ \text{o} \ (3,2,1,4) $$

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.

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

FacebookTwitterLinkedinLinkedin
knowledge base

Groups