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

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## 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.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $12$