Lagrange's Theorem (Group Theory)/Examples/Order of Union of Subgroups of Order 16

Examples of Use of Lagrange's Theorem
Let $G$ be a group whose identity is $e$.

Let $H$ and $K$ be subgroups of $G$ such that:
 * $\order H = \order K = 16$
 * $H \ne K$

where $\order {\, \cdot \,}$ denotes the order of the subgroup.

Then:
 * $24 \le \order {H \cup K} \le 31$

Proof
As $H$ and $K$ are subgroups of $G$, they share at least $e$.

That is, $\order {H \cap K} \ge 1$.

On the other hand, we have that $H \ne K$.

Thus $\order {H \cap K} < 16$.

But from Intersection of Subgroups is Subgroup, $H \cap K$ is a subgroup of both $H$ and $K$.

From Lagrange's Theorem, that means:
 * $\order {H \cap K} \divides 16$

and so:
 * $\order {H \cap K} \le 8$

From Cardinality of Set Union:


 * $\order {H \cup K} = \order H + \order K = \order {H \cap K}$

and so:
 * $\order {H \cup K} = 16 + 16 = \order {H \cap K} = 32 - \order {H \cap K}$

As $\order {H \cap K}$ lies within the range of $1$ to $8$, the result follows.