Groups: Axioms

This week I wanted to discuss a topic that I’ve been learning about in lectures: groups. In this post I will highlight the axioms that a set of numbers must satisfy in order for it to be a group, i.e. the definition of a group.

So, a group (G,) (where  is the operation) must satisfy the following conditions:

  • Closure: if a and b are two elements in G, then a∘b is also an element in G

∀ a,∈ a∈ G   

  • Associativity: the defined operation is associative

∀ a,b,∈ : a(bc(ab)c

  • Identity: there is an identity element e which is in G

∃ ∈ ∀ ∈ Geae

  • Inverse: each element a in G must have an inverse b that is also in G

∀ ∈ ∃ ∈ ab∘a

A group is abelian if it is commutative, i.e.∀ a,∈ G: a•b = b•a.

EXAMPLE: Prove (Z,+) is a group.

To prove that something is a group, we must go through the group axioms and check that all of them hold.

  • For any two integers a, b, the sum of these integers a+b is also an integer. Therefore the set is closed under addition.
  • For all integers a, b, c, (a+b) + c = a + (b+c), hence associativity is satisfied.
  • If a is an integer, then a + 0 = a = 0 + a, so the identity element is 0, which is also an integer. Therefore, this set contains the identity element 0.
  • For every integer a, there is an integer b such that a + b = 0 = b + a, as this occurs when b = -a. Hence, each number in the set has an inverse element which is also in the set of integers.

Concluding, it follows that the set of integers is a group under addition. Note that this group is also abelian as a + b = b + a for all integers a, b.

Sources: 1 | 2 | 3 | 4

Thursday’s post will be about subgroups! M x


3 thoughts on “Groups: Axioms”

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s