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 H of a group G is a subgroup if and only if

(i) \left( {a \in H,\,b \in H} \right) \Rightarrow a \circ b \in H and

(ii) a \in H \Rightarrow {a^{ - 1}} \in H

Proof: Suppose H is a subgroup of G, then H must be closed with respect to composition  \circ in G, i.e. a \in H,\,b \in H \Rightarrow a \circ b \in H.

Let a \in H and {a^{ - 1}} be the inverse of a in G. Then the inverse of a in H is also {a^{ - 1}}. As H itself is a group, each element of H will possess inverse in it, i.e. a \in H \Rightarrow {a^{ - 1}} \in H.

Thus the condition is necessary. Now let us examine the sufficiency of the condition.

(i) Closure Axiom: a \in H,\,b \in H \Rightarrow a \circ b \in H. Hence the closure axiom is satisfied with respect to the operation  \circ .

(ii) Associative Axiom: Since the elements of H are also the elements of G, the composition is associative in H also.

(iii) Existence of Identity: The identity of the subgroup is the same as the identity of the group because a \in H,\,{a^{ - 1}} \in H \Rightarrow a \circ {a^{ - 1}} \in H \Rightarrow e \in H. The identity e is an element of H.

(iv) Existence of Inverse: Since a \in H \Rightarrow {a^{ - 1}} \in H,\,\,\,\forall a \in H. Therefore each element of H possesses an inverse.

The H itself is a group for the composition  \circ in G. Hence H is a subgroup.

Theorem 2: A necessary and sufficient condition for a non-empty subset H of a group G to be a subgroup is that a \in H,\,b \in H \Rightarrow a \circ {b^{ - 1}} \in H where {b^{ - 1}} is the inverse of b in G.

Proof: The condition is necessary. Suppose H is a subgroup of G and let a \in H,\,b \in H.

Now each element of H must possess an inverse because H itself is a group.

 b \in H \Rightarrow {b^{ - 1}} \in H

Also H is closed under the composition  \circ in G. Therefore

 a \in H,\,{b^{ - 1}} \in H \Rightarrow a \circ {b^{ - 1}} \in H

The condition is sufficient. If it is given that a \in H,\,{b^{ - 1}} \in H \Rightarrow a \circ {b^{ - 1}} \in H then we have to prove that H is a subgroup.

(i) Closure Property: Let a,b \in H then b \in H \Rightarrow {b^{ - 1}} \in H (as shown above). Therefore by the given condition:

\begin{gathered} a \in H,\,{b^{ - 1}} \in H \Rightarrow a \circ {\left( {{b^{ - 1}}} \right)^{ - 1}} \in H \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a \circ b \in H \\ \end{gathered}

Thus H is closed with respect to the composition  \circ in G.

(ii) Associative Property: Since the elements of H are also the elements of G, the composition is associative in H.

(iii) Existence of Identity: Since

a \in H,\,{a^{ - 1}} \in H \Rightarrow a \circ {a^{ - 1}} \in H \Rightarrow e \in H

(iv) Existence of Inverse: Let a \in H, then

e \in H,\,a \in H \Rightarrow e \circ {a^{ - 1}} \in H \Rightarrow {a^{ - 1}} \in H

Then each element of H possesses an inverse.

Hence H itself is a group for the composition  \circ in group G.