Intersection of Coprime Cyclic Subgroups is Trivial

Theorem
Let $G$ be a group whose identity is $e$.

Let $x, y \in G$ such that:
 * $\order x \perp \order y$

where:
 * $\order x, \order y$ denotes the orders of $x$ and $y$ in $G$ respectively
 * $\perp$ denotes the coprimality relation.

Then:
 * $\gen x \cap \gen y = \set e$

where $\gen x, \gen y$ denotes the subgroup of $G$ generated by $x$ and $y$ in $G$ respectively.

Proof
From Order of Cyclic Group equals Order of Generator:
 * $\order x = \order {\gen x}$

and:
 * $\order y = \order {\gen y}$

where $\order {\gen x}, \order {\gen y}$ denote the orders of $\gen x$ and $\gen y$ respectively.

From Intersection of Subgroups is Subgroup:
 * $\gen x \cap \gen y$ is a subgroup of both $\gen x$ and $\gen y$.

From Lagrange's Theorem:


 * $\order {\gen x \cap \gen y} \divides \order {\gen x}$

and:
 * $\order {\gen x \cap \gen y} \divides \order {\gen y}$

But as $\order {\gen x} \perp \order {\gen y}$:
 * $\order {\gen x \cap \gen y} = 1$

Hence the result by definition of trivial subgroup.