The necessary and sufficient conditions for a subset of a group to be a subgroup are stated in the following two theorems.
Theorem 1: A subset of a group is a subgroup if and only if
Proof: Suppose is a subgroup of , then must be closed with respect to composition in , i.e. .
Let and be the inverse of in . Then the inverse of in is also . As itself is a group, each element of will possess inverse in it, i.e. .
Thus the condition is necessary. Now let us examine the sufficiency of the condition.
(i) Closure Axiom: . Hence the closure axiom is satisfied with respect to the operation .
(ii) Associative Axiom: Since the elements of are also the elements of , the composition is associative in also.
(iii) Existence of Identity: The identity of the subgroup is the same as the identity of the group because . The identity is an element of .
(iv) Existence of Inverse: Since . Therefore each element of possesses an inverse.
The itself is a group for the composition in . Hence is a subgroup.
Theorem 2: A necessary and sufficient condition for a non-empty subset of a group to be a subgroup is that where is the inverse of in .
Proof: The condition is necessary. Suppose is a subgroup of and let .
Now each element of must possess an inverse because itself is a group.
Also is closed under the composition in . Therefore
The condition is sufficient. If it is given that then we have to prove that is a subgroup.
(i) Closure Property: Let then (as shown above). Therefore by the given condition:
Thus is closed with respect to the composition in .
(ii) Associative Property: Since the elements of are also the elements of , the composition is associative in .
(iii) Existence of Identity: Since
(iv) Existence of Inverse: Let , then
Then each element of possesses an inverse.
Hence itself is a group for the composition in group .