# 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 ${H_2} \subset {H_1}$.