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 pm
In 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 am
the proper order of the proof is (iii) (iv) (i) (ii)
Naveena Chandran
April 12 @ 2:54 pm
The condition is sufficient. It is given that a ∈ H,b ∈ H⇒a*inverse (b) ∈ H
Claim: H is a group (subgroup)
i). Taking b =a, then a*inverse (a) = e ∈ H
ii). For a in H, e*inverse (a) = inverse (a) ∈ H
iii). For b in H, by (ii), inverse (b) is in H and so a*inverse (inverse (b)) ∈ H
Also, note that inverse (inverse (b)) = b in G.
Thus, for a, b in H, a*b = a*inverse (inverse (b)) ∈ H
(iv) The associativity in H is inherited from that in G
Hence, all four axioms are satisfied for H.