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Let [ilmath]G[/ilmath] be a set and a binary operation (a function) [ilmath]*:G\times G\rightarrow G[/ilmath], with the following properties[1]:

  1. [ilmath]\forall g,h,k\in G[(g*h)*k=g*(h*k)][/ilmath] - is associative
  2. [ilmath]\exists e\in G\ \forall g\in G[g*e=e*g=g][/ilmath] - there exists an identity element of [ilmath]G[/ilmath][Note 1]
  3. [ilmath]\forall g\in G\exists h\in G[g*h=h*g=e][/ilmath] - for each element there exists an inverse element in [ilmath]G[/ilmath][Note 2]

If [ilmath]*[/ilmath] satisfies these 3 properties than the tuple, [ilmath](G,*:G\times G\rightarrow G)[/ilmath] - or just [ilmath](G,*)[/ilmath] as mathematicians are lazy, is called a group.
Claims: (see below for proof)

  • Claim 1: The identity element is unique
  • Claim 2: The inverse element of an element is unique

Abelian group

If, additionally, a group [ilmath](G,*)[/ilmath] satisfies and additional property:

  1. [ilmath]\forall g,h\in G[g*h=h*g][/ilmath] - the operation is commutative

then we call the group an Abelian group or commutative group

Terminology and notations

Proof of claims

Group/Claim 1: The identity element of a group is unique Group/Claim 2: The inverse element for each element of a group is unique


  1. At this point we do not know that the identity element is unique, there could be more than one such [ilmath]e[/ilmath] - but one exists. In fact it is unique, as we will see later
  2. Again, we do not know there is a unique inverse, or for which of the [ilmath]e[/ilmath] elements the equality refers to (if there are even more than one).
    • There is actually only one identity element, only one [ilmath]e\in G[/ilmath] and also only one inverse for each element, but this requires proof.


  1. Fundamentals of Abstract Algebra - Neal H. McCoy