Editing Group (mathematics) (section)

== Elementary consequences of the group axioms ==
Basic facts about all groups that can be obtained directly from the group axioms are commonly subsumed under ''elementary group theory''.{{sfn|Ledermann|1953|loc=§1.2|pp=4–5}} For example, [[Mathematical induction|repeated]] applications of the associativity axiom show that the unambiguity of
<math display=block>a\cdot b\cdot c=(a\cdot b)\cdot c=a\cdot(b\cdot c)</math>
generalizes to more than three factors. Because this implies that [[Bracket#Parentheses in mathematics|parentheses]] can be inserted anywhere within such a series of terms, parentheses are usually omitted.{{sfn|Ledermann|1973|loc=§I.1|p=3}}

=== Uniqueness of identity element ===
The group axioms imply that the identity element is unique; that is, there exists only one identity element: any two identity elements <math>e</math> and <math>f</math> of a group are equal, because the group axioms imply {{tmath|1= e=e\cdot f=f }}. It is thus customary to speak of ''the'' identity element of the group.{{sfn|Lang|2005|loc=§II.1|p=17}}

=== Uniqueness of inverses ===
The group axioms also imply that the inverse of each element is unique. Let a group element <math>a</math> have both <math>b</math> and <math>c</math> as inverses. Then
: <math>\begin{align}
b &=  b\cdot e 
    && \text{(}e \text { is the identity element)}\\
  &=  b\cdot (a \cdot c) 
    && \text{(}c \text { and } a \text{ are inverses of each other)}\\
  &=  (b\cdot a) \cdot c 
    && \text{(associativity)}\\
  &=  e \cdot c 
    && \text{(}b \text { is an inverse of } a\text{)}\\
  &=  c 
    && \text{(}e \text { is the identity element and } b=c\text{)}
\end{align}</math>

Therefore, it is customary to speak of ''the'' inverse of an element.{{sfn|Lang|2005|loc=§II.1|p=17}}

=== <span id="translation"></span> Division ===
Given elements <math>a</math> and <math>b</math> of a group {{tmath|1= G }}, there is a unique solution <math>x</math> in <math>G</math> to the equation {{tmath|1= a\cdot x=b }}, namely {{tmath|1= a^{-1}\cdot b }}.{{efn|One usually avoids using fraction notation <!--use {{math}}, since <math> in footnotes is unreadable on mobile devices-->{{math|{{sfrac|''b''|''a''}}}} unless {{math|''G''}} is abelian, because of the ambiguity of whether it means {{math|''a''<sup>−1</sup> ⋅ ''b''}} or {{math|''b'' ⋅ ''a''<sup>−1</sup>}}.)}}{{sfn|Artin|2018|p=40}} It follows that for each <math>a</math> in {{tmath|1= G }}, the function <math>G\to G</math> that maps each <math>x</math> to <math>a\cdot x</math> is a [[bijection]]; it is called ''left multiplication'' by <math>a</math> or ''left translation'' by {{tmath|1= a }}.

Similarly, given <math>a</math> and {{tmath|1= b }}, the unique solution to <math>x\cdot a=b</math> is {{tmath|1= b\cdot a^{-1} }}. For each {{tmath|1= a }}, the function <math>G\to G</math> that maps each <math>x</math> to <math>x\cdot a</math> is a bijection called ''right multiplication'' by <math>a</math> or ''right translation'' by {{tmath|1= a }}.

=== Equivalent definition with relaxed axioms ===
The group axioms for identity and inverses may be "weakened" to assert only the existence of a [[left identity]] and [[left inverse element|left inverse]]s. From these ''one-sided axioms'', one can prove that the left identity is also a right identity and a left inverse is also a right inverse for the same element. Since they define exactly the same structures as groups, collectively the axioms are not weaker.{{sfn|Lang|2002|loc=§I.2|p=7}}

In particular, assuming associativity and the existence of a left identity <math>e</math> (that is, {{tmath|1= e\cdot f=f }}) and a left inverse <math>f^{-1}</math> for each element <math>f</math> (that is, {{tmath|1= f^{-1}\cdot f=e }}), it follows that every left inverse is also a right inverse of the same element as follows.{{sfn|Lang|2002|loc=§I.2|p=7}}
Indeed, one has
: <math>\begin{align}
f \cdot f^{-1}
  &=e \cdot (f \cdot f^{-1}) 
       && \text{(left identity)}\\
  &=((f^{-1})^{-1} \cdot f^{-1}) \cdot (f \cdot f^{-1})
       && \text{(left inverse)}\\
  &=(f^{-1})^{-1} \cdot ((f^{-1} \cdot f) \cdot f^{-1})
       && \text{(associativity)}\\
  &=(f^{-1})^{-1} \cdot (e \cdot f^{-1})
       && \text{(left inverse)}\\
  &=(f^{-1})^{-1} \cdot f^{-1}
       && \text{(left identity)}\\
  &=e
       && \text{(left inverse)}
\end{align}</math>

Similarly, the left identity is also a right identity:{{sfn|Lang|2002|loc=§I.2|p=7}}
: <math>\begin{align}
f\cdot e 
  &= f \cdot ( f^{-1} \cdot f)
       && \text{(left inverse)}\\
  &= (f \cdot  f^{-1}) \cdot f
       && \text{(associativity)}\\
  &= e \cdot f
       && \text{(right inverse)}\\
  &= f
       && \text{(left identity)}
\end{align}</math>

These results do not hold if any of these axioms (associativity, existence of left identity and existence of left inverse) is removed. For a structure with a looser definition (like a [[semigroup]]) one may have, for example, that a left identity is not necessarily a right identity.

The same result can be obtained by only assuming the existence of a right identity and a right inverse.

However, only assuming the existence of a ''left'' identity and a ''right'' inverse (or vice versa) is not sufficient to define a group. For example, consider the set <math>G = \{ e,f \}</math> with the operator <math>\cdot</math> satisfying <math>e \cdot e = f \cdot e = e</math> and {{tmath|1= e \cdot f = f \cdot f = f }}. This structure does have a left identity (namely, {{tmath|1= e }}), and each element has a right inverse (which is <math>e</math> for both elements). Furthermore, this operation is associative (since the product of any number of elements is always equal to the rightmost element in that product, regardless of the order in which these operations are applied). However, <math>( G , \cdot )</math> is not a group, since it lacks a right identity.