# Necessary and Sufficient Condition for a Subgroup

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

**(i)** and

**(ii)**

**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 .

john

July 31@ 9:17 pmIn proving sufficiency, you don't seem to have justified the (true) statement that 'b' in H implies its inverse is in H.