## Section 14.2 Examples of Groups

In order to determine whether or not a set with a binary operation defined on the set forms a group, we must investigate whether or not each of the properties in Item 1 to Item 4 from Definition 14.1.2 are met. If all of the properties are met, we conclude that the set with the operation defined on it forms a group. If even one of the properties is not met, we conclude that the set with the operation defined on it does not form a group.

In the video in Figure 14.2.1 we give first examples of groups. Below we go through such examples in detail.

We now piece together information developed in the examples of Chapter 13 to check whether these binary operations give us groups.

### Problem 14.2.2. Is \((\Z,+)\) a group ?

Is the set \(\Z\) of integers with addition a commutative group?

As the sum of two integers is an integer, addition is a binary operation on the set of integers \(\Z\text{.}\) We check whether the set of integers \(\Z\) with the operation addition (\(+\)) fulfills the properties Item 1 to Item 4 from Definition 14.1.2:

*Identity:*For all \(a\in\Z\) we have \(a+0=a\) and \(0+a=a\text{,}\) hence the integer 0 is the identity element with respect to addition. (Compare Example 13.3.4 (a).)*Inverses:*For all \(a\in \Z\) we have \(a+(-a)=0\) and \((-a)+a=0\text{,}\) hence the inverse of \(a\) with respect to addition is the integer \(-a\text{.}\) So every integer has an inverse. (Compare Example 13.4.8 (a).)*Associativity*: Addition of integers is associative. (See Example 13.2.4 (a).)*Commutativity:*Addition of integers is commutative. (See Example 13.5.4 (a).)

Thus \((\Z,+)\) is a commutative group.

### Problem 14.2.3. Is \((\N,\cdot)\) a group ?

Is the set \(\N\) of natural numbers with multiplication a commutative group?

By Example 13.4.8 (b), there is a natural number that does not have a multiplicative inverse. So, property Item 2 is not fulfilled, and we conclude that the set of natural numbers with multiplication is not a commutative group. The properties Item 1, Item 3, and Item 4 are met by part (b) of Example 13.3.4, Example 13.2.4, and Example 13.5.4, respectively.

### Problem 14.2.4. Is \((T,\star)\) a group ?

Is the set \(T=\{\Tx,\Ty,\Tz\}\) with the binary operation \(\star:T\to T\) that is given by the table in Example 13.1.4 a commutative group?

We consider the properties Item 1 to Item 4 from Definition 14.1.2:

*Identity:*The identity element is \(\Ty\text{.}\) (See Example 13.3.6 and Example 13.3.7.)*Inverses:*The inverse of \(\Tx\) is \(\Tx^{-1\star} = \Tz\text{,}\) the inverse of \(\Ty\) is \(\Ty^{-1\star} = \Ty\text{,}\) and the inverse of \(\Tz\) is \(\Tz^{-1\star} = \Tx\text{,}\) so every element of \(T\) has an inverse. (See Example 13.4.10 and Example 13.4.11.)*Associativity*: \(\star\) is associative. (See Example 13.2.5.)*Commutativity:*\(\star\) is commutative. (See Example 13.5.5 and Example 13.5.6.)

Thus \((T,\star)\) is a commutative group.

Similarly it follows from Example 13.3.8, Example 13.4.12, Example 13.2.6, and Example 13.5.7 that \((\Z_5,\oplus)\) where \(\Z_5=\{0,1,2,3,4\}\) and \(\oplus:\Z_5\times\Z_5\to\Z_5\) is defined by \(a\oplus b=(a+b)\fmod 5\) is a group.

So far we have considered all of the sets and operations that were given in Chapter 13. Now it is time to investigate further examples. A commutative group must contain at least one element, namely the identity. The following problem demonstrates that there are commutative groups with just one element.

### Problem 14.2.5. Is \((\{1\}, \cdot)\) a group ?

Is \((\{1\}, \cdot)\text{,}\) where \(\cdot\) is multiplication, a commutative group?

We consider the properties Item 1 to Item 4 from Definition 14.1.2. Since \(1 \cdot 1 = 1 \in \{1\}\) we have that \(\cdot\) is a binary operation on \(\{1\}\text{.}\)

*Identity:*The identity element is \(1\text{.}\)*Inverses:*The inverse of \(1\) is \(1^{-1\cdot} = 1\text{,}\) so every element of \(\{1\}\) has an inverse.*Associativity*: \(1\cdot(1\cdot 1)=1\cdot 1=1\) and \((1\cdot 1)\cdot 1=1\cdot 1=1\text{.}\) Since \(1 = 1\text{,}\) we have that \(\cdot\) is associative.*Commutativity:*\(1\cdot 1=1\cdot 1\text{,}\) so \(\cdot\) is commutative.

Thus \(\left(\{1\},\cdot\right)\) is a commutative group.

We conclude by taking one more look at a set with an operation defined by a table.

### Example 14.2.6. A group given by an operation table.

Let \(S=\{\So,\Sno,\Si,\Sni\}\text{,}\) and let the operation \(\diamond:S\diamond S\to S\) be given by the operation table:

\(\diamond\) | \(\So\) | \(\Sno\) | \(\Si\) | \(\Sni\) |

\(\So\) | \(\So\) | \(\Sno\) | \(\Si\) | \(\Sni\) |

\(\Sno\) | \(\Sno\) | \(\So\) | \(\Sni\) | \(\Si\) |

\(\Si\) | \(\Si\) | \(\Sni\) | \(\Sno\) | \(\So\) |

\(\Sni\) | \(\Sni\) | \(\Si\) | \(\So\) | \(\Sno\) |

Each entry in the table is an element in \(S\text{,}\) so \(\diamond\) is a binary operation on \(S\text{.}\) We show that \((S,\diamond)\) is a commutative group by verifying that the properties Item 1 to Item 4 from Definition 14.1.2 hold.

*Identity:*The row corresponding to the element \(\So\) matches the header row at the top of the table and the column corresponding to the element \(\So\) matches the header column on the left side of the table. Thus the identity element is \(\So\text{.}\)*Inverses:*Since \(\So\) is the identity element, we begin by locating all of the places \(\So\) appears in the table. For each table entry that is \(\So\text{,}\) we trace back to the header column on the left side of the table and the header row on the top of the table. Since \(\So \diamond \So = \So\text{,}\) we conclude that \(\So\) is its own inverse, and since \((\Sno) \diamond (\Sno) = \So\text{,}\) we conclude that \(\Sno\) is also its own inverse. Finally, since \(\Si \diamond (\Sni) = \So\) and \((\Sni) \diamond \Si = \So\text{,}\) we conclude that \(\Si\) and \(\Sni\) are inverses of each other. Thus every element of \(S\) has an inverse.*Associativity*: Exhausting all 64 possibilities, we would be able to see that the operation \(\diamond\) is associative.*Commutativity:*The table is symmetric about the diagonal from the operation symbol in the top left corner of the table to the bottom right corner of the table. Thus \(\diamond\) is commutative.

Thus \((S,\diamond)\) is a commutative group.

Im Checkpoint 14.2.7 determine whether the given set and binary operation yields a group.

### Checkpoint 14.2.7. Is this a group ?

Let the binary operation \(\diamond\) on the set \(F=\lbrace\) a, b, c, d, x, y, z\(\rbrace\) be defined by:

\(\diamond\) | a | b | c | d | x | y | z |

a | a | z | y | x | d | c | b |

b | b | a | z | y | x | d | c |

c | c | b | a | z | y | x | d |

d | d | c | b | a | z | y | x |

x | x | d | c | b | a | z | y |

y | y | x | d | c | b | a | z |

z | z | y | x | d | c | b | a |

Complete the following:

(1) In the set \(F\) with respect to \(\diamond\)

select

the identity element is a

the identity element is b

the identity element is c

the identity element is d

the identity element is x

the identity element is y

the identity element is z

there is no identity element

(2) In the set \(F\)

select

each element has an inverse

at least one element does not have an inverse

there is no identity, so inverses are not defined

(4) The operation \(\diamond\) is not associative.

(e) The operation \(\diamond\) is

select

commutative

not commutative

Conclude whether \((F,\diamond)\) is a commutative group:

The set \(F\) with the operation \(\diamond\) is

select

a commutative group

not a commutative group

With the knowledge aquired above determine whether the sets and binary operations given in Checkpoint 14.2.8 are groups.

### Checkpoint 14.2.8. Are these groups ?

Are these sets with the given operations commutative groups ?

If not, indicate the reason. If several reasons apply, select the first reason that applies.

(\(\lbrace 0\rbrace\text{,}\)+)

(\(\lbrace-1,1\rbrace\text{,}\)*) where \(a\)*\(b:=a\cdot b\)

(\(\mathbb{N}\text{,}\)+)

(\(\lbrace1\rbrace\text{,}\)*) where \(a\)*\(b := a\cdot b\)

(\(\mathbb{Z}_{3}\text{,}\)*) where \(a\)*\(b := (a\cdot b)\bmod 3\)