# 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 $G$.