Intersection of Coprime Cyclic Subgroups is Trivial
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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.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $17$