# 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 $$G$$.

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.

jinsong li

May 2@ 6:38 amthe proper order of the proof is (iii) (iv) (i) (ii)