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 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 belongs 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} are two subgroups of a groupG, 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}.