Learn about group axioms and finite groups in this lesson.

Binary operators


A binary operator on a nonempty set GG is a function ∗*, which combines two given elements a,b∈Ga, b \in G and maps them to a new element ∗(a,b)*(a, b), to say i.e.,

∗:G×G→G,(a,b)↦∗(a,b)*: G \times G \rightarrow G, \quad(a, b) \mapsto *(a, b)

Instead of writing ∗(a,b)*(a, b), we write a∗ba * b from now on.


The addition ++ is a binary operation defined on the integers Z.\mathbb{Z} . Instead of writing +(5,7)=12+(5,7)=12, we write 5+7=125+7=12.

Note: For any given a,b∈Ga, b \in G, it follows that a∗b∈Ga * b \in G, according to the definition given above. Thus, we say that the set GG is closed with respect to (∗)(*).


We consider the following examples of closure:

  1. The set of integers Z\mathbb{Z} is closed with respect to the binary operation ++ of addition, meaning that the sum of two integers is also an integer.
  2. The set 2Z2 \mathbb{Z} of even integers is closed under addition because the sum of two even integers gives an even integer since 2a+2b=2(a+b)∈2Z2 a+2 b=2(a+b) \in 2 \mathbb{Z}. Conversely, the set Z\2Z\mathbb{Z} \backslash 2 \mathbb{Z} of odd integers is not closed under addition because the addition of two odd integers always yields an even integer.

The group axioms


A set of elements GG together with a binary operation ∗* is called a group if the following axioms are satisfied:

  • Closure property: The group operation ∗* is closed, i.e., for all a,b∈Ga, b \in G, it holds that a∗b=a * b= c∈Gc \in G.

  • Associativity: The group operation * is associative, i.e., a∗(b∗c)=(a∗b)∗ca *(b * c)=(a * b) * c for all a,b,c∈Ga, b, c \in G.

    1. Identity property: There’s an identity or neutral element e∈Ge \in G, such that, e∗a=a∗e=e * a=a * e= aa, for all a∈Ga \in G.

    2. Inverse property: For each a∈Ga \in G, there exists an inverse element of aa, namely a−1a^{-1}, such that a−1∗a=a∗a−1=ea^{-1} * a=a * a^{-1}=e.

Furthermore, a group GG is said to be abelian (or commutative) if a∗b=b∗aa * b=b * a, for all a,b∈Ga, b \in G.

Note: We usually write (G,∗)(G, *) to GG in order to make clear that the group GG takes the operation ∗* as a basis. If (G,∗)(G, *) forms an additive group, we write a∗b=a * b= a+ba+b, whereas the neutral element is written as e=0e=0, and the inverse element to aa is given by −a.-a . In case the group is multiplicative, we write a∗b=a⋅b=aba * b=a \cdot b=a b and denote the neutral element by e=1e=1 and the inverse by a−1=1aa^{-1}=\frac{1}{a}.


  1. (N0,+)\left(\mathbb{N}_{0},+\right) doesn’t form a group since there’s no inverse element to aa, such that a+(−a)=0a+(-a)=0.

  2. (Z,+)(\mathbb{Z},+) with the neutral element 0 and the inverse element −a-a forms an abelian group because it holds that 0+a=a+0=a0+a=a+0=a for all a∈Za \in \mathbb{Z} and (−a)+a=a+(−a)=0(-a)+a=a+(-a)=0 for all a∈Za \in \mathbb{Z}.

  3. (Z,+),(R,+)(\mathbb{Z},+),(\mathbb{R},+), and (C,+)(\mathbb{C},+) form abelian groups with e=0e=0 and a−1=a^{-1}= −a-a.

  4. (Q,⋅)(\mathbb{Q}, \cdot) and (R,⋅)(\mathbb{R}, \cdot) are not groups because there’s no inverse element for 00.

  5. We denote Q∗=Q\{0}\mathbb{Q}^{*}=\mathbb{Q} \backslash\{0\} and R∗=R\{0}\mathbb{R}^{*}=\mathbb{R} \backslash\{0\}. Then, (Q∗,⋅)\left(\mathbb{Q}^{*}, \cdot\right) and (R∗,⋅)\left(\mathbb{R}^{*}, \cdot\right) are abelian groups with e=1e=1 and a−1=1a.a^{-1}=\frac{1}{a} . Note that Z∗=Z\{0}\mathbb{Z}^{*}=\mathbb{Z} \backslash\{0\} is not a group, because there is no inverse element 1a\frac{1}{a} for any element a.a .

Proposition 1:

Let n∈Nn \in \mathbb{N} with n≥2.n \geq 2 . Then, (Zn,⊕)(\left(\mathbb{Z}_{n}, \oplus\right)( with ⊕\oplus of the definition: Addition and multiplication modulo :Addition_and_Multiplication_Modulo ) is an abelian group.


Let a,b,c∈Zn.a, b, c \in \mathbb{Z}_{n} . We show that the group axioms are satisfied:

  • G0: The group operation ⊕\oplus is closed because of the operator’s definition: addition and multiplication modulo:Addition_and_Multiplication_Modulo

  • G1: Because of the associativity of (Z,+)(\mathbb{Z},+), we conclude that

aˉ⊕(bˉ⊕cˉ)=aˉ⊕b+c‾=a+(b+c)‾=(a+b)+c‾=a+b‾⊕cˉ=(aˉ⊕bˉ)⊕cˉ,\begin{aligned} \bar{a} \oplus(\bar{b} \oplus \bar{c}) &=\bar{a} \oplus \overline{b+c}=\overline{a+(b+c)}=\overline{(a+b)+c}=\overline{a+b} \oplus \bar{c} \\ &=(\bar{a} \oplus \bar{b}) \oplus \bar{c}, \end{aligned}

which shows that ⊕\oplus is associative.

  • G2:
  1. The element 0∈Zn0 \in \mathbb{Z}_{n} is the identity element because

0‾⊕aˉ=0+a‾=aˉ\overline{0} \oplus \bar{a}=\overline{0+a}=\bar{a}


aˉ⊕0‾=a+0‾=aˉ\bar{a} \oplus \overline{0}=\overline{a+0}=\bar{a}

for every aˉ∈Zn\bar{a} \in \mathbb{Z}_{n}.

  1. There’s an inverse for each element aˉ∈Zn\bar{a} \in \mathbb{Z}_{n}, because

aˉ⊕n−a‾=a+(n−a)‾=nˉ=0‾\bar{a} \oplus \overline{n-a}=\overline{a+(n-a)}=\bar{n}=\overline{0}

for every aˉ∈Zn\bar{a} \in \mathbb{Z}_{n}.

Furthermore, the group (Zn,⊕)\left(\mathbb{Z}_{n}, \oplus\right) is commutative because

aˉ⊕bˉ=a+b‾=b+a‾=bˉ⊕aˉ\bar{a} \oplus \bar{b}=\overline{a+b}=\overline{b+a}=\bar{b} \oplus \bar{a}

for every aˉ,bˉ∈Zn\bar{a}, \bar{b} \in \mathbb{Z}_{n} since (Z,+)(\mathbb{Z},+) is also commutative.

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