Show that the set of all integers …-4, -3, -2, -1, 0, 1, 2, 3, 4, ... is an infinite Abelian group with respect to the operation of addition of integers.
Let us test all the group axioms for an Abelian group.
(G1) Closure Axiom: We know that the sum of any two integers is also an integer, i.e., for all , . Thus is closed with respect to addition.
(G2) Associative Axiom: Since the addition of integers is associative, the associative axiom is satisfied, i.e., for such that
(G3) Existence of Identity: We know that is the additive identity and , i.e.,
Hence, additive identity exists.
(G4) Existence of Inverse: If , then . Also,
Thus, every integer possesses additive inverse. Therefore is a group with respect to addition.
Since the addition of integers is a commutative operation, therefore
Hence is an Abelian group. Also, contains an infinite number of elements.
Therefore is an Abelian group of infinite order.
Show that the set of all non-zero rational numbers with respect to the operation of multiplication is a group.
Let the given set be denoted by . Then by group axioms, we have
(G1) We know that the product of two non-zero rational numbers is also a non-zero rational number. Therefore is closed with respect to multiplication. Hence, the closure axiom is satisfied.
(G2) We know for rational numbers:
Hence, the associative axiom is satisfied.
(G3) Since the multiplicative identity is a rational number, hence the identity axiom is satisfied.
(G4) If , then obviously, . Also
so that is the multiplicative inverse of . Thus the inverse axiom is also satisfied. Hence is a group with respect to multiplication.
Show that , the set of all non-zero complex numbers is a multiplicative group.
Let . Here is the set of all real numbers and .
(G1) Closure Axiom: If and , then by the definition of multiplication of complex numbers
Since , for . Therefore, is closed under multiplication.
(G2) Associative Axiom:
(G3) Identity Axiom: is the identity in .
(G4) Inverse Axiom: Let , then
Hence is a multiplicative group.