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 Hwill 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 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 Hpossesses inverse.
The Hitself 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 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 inverse.
Hence H itself is a group for the composition  \circ in group G.

Comments

comments