Properties of Subgroups

Theorem 1: The intersection of two subgroups of a group $$G$$ is a subgroup of $$G$$.

Proof: Let $${H_1}$$ and $${H_2}$$ be any two subgroups of $$G$$. Then $${H_1} \cap {H_2} \ne \phi $$ because at least the identity element $$e$$ is common in both $${H_1}$$ and $${H_2}$$.

Now to prove that $${H_1} \cap {H_2}$$ is a subgroup of $$G$$, it is sufficient to show that $$a \in {H_1} \cap {H_2}$$, $$b \in {H_1} \cap {H_2}$$$$ \Rightarrow a \circ {b^{ – 1}} \in {H_1} \cap {H_2}$$, $$ \circ $$ being a composition in $$G$$.

Since $$a \in {H_1} \cap {H_2}$$$$ \Rightarrow a \in {H_1}$$ and $$b \in {H_2}$$ and $$b \in {H_1} \cap {H_2} \Rightarrow {H_1}$$ and $$b \in {H_2}$$ and $${H_1},{H_2}$$ are subgroups of $$G$$ we see that $$a \in {H_1},b \in {H_1}$$$$ \Rightarrow a \circ {b^{ – 1}} \in {H_1}$$ and similarly $$a \in {H_2},b \in {H_2}$$$$ \Rightarrow a \circ {b^{ – 1}} \in {H_2}$$.

Thus, $$a \circ {b^{ – 1}} \in {H_1},a \circ {b^{ – 1}} \in {H_2} \Rightarrow a \circ {b^{ – 1}} \in {H_1} \cap {H_2}$$, which establishes that $${H_1} \cap {H_2}$$ is a subgroup of $$G$$.
 

Theorem 2: The union of two subgroups is not necessarily a subgroup.

Proof: For example, let $$O$$ be the additive group of integers, and let
\[{H_1} = \left\{ {0,\, \pm 2,\, \pm 4,\, \pm 6, \ldots } \right\}\]
\[{H_2} = \left\{ {0,\, \pm 3,\, \pm 6,\, \pm 9, \ldots } \right\}\]

Then $${H_1},\,{H_2}$$ are subgroups of $$G$$, but  $${H_1} \cup {H_2} = \left\{ {0,\, \pm 2,\, \pm 3,\, \pm 4,\, \pm 6, \ldots } \right\}$$, which is not a group. It is evident that the closure property is not satisfied. For, 2 + 3 =5, which does not belong to$${H_1} \cup {H_2}$$.

The set $${H_1} \cap {H_2} = \left\{ {0,\, \pm 6,\, \pm 12,\, \ldots } \right\}$$ is certainly a group.

Theorem 3: The union of two subgroups is a subgroup if and only if one is contained in the other.
Let $${H_1},{H_2}$$ be two subgroups of a group$$G$$, then $${H_1} \cup {H_2}$$ is a subgroup if and only if either $${H_1} \subset {H_2}$$ or $${H_2} \subset {H_1}$$.