Intersection of Subgroups of Prime Order

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Theorem

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

Let $H$ and $K$ be subsets of $G$ such that:

$\order H = \order K = p$
$H \ne K$
$p$ is prime.


Then:

$H \cap K = \set e$


That is, the intersection of two unequal subgroups of a group, both of whose order is the same prime, consists solely of the identity.


Proof

From Intersection of Subgroups is Subgroup:

$H \cap K \le G$

and:

$H \cap K \le H$

where $\le$ denotes subgrouphood.


So:

\(\ds H \cap K\) \(\le\) \(\ds H\) Intersection of Subgroups is Subgroup
\(\ds \leadsto \ \ \) \(\ds \order {H \cap K}\) \(\divides\) \(\ds \order H\) Lagrange's Theorem
\(\ds \leadsto \ \ \) \(\ds \order {H \cap K}\) \(\divides\) \(\ds p\)
\(\ds \leadsto \ \ \) \(\ds \order {H \cap K}\) \(=\) \(\ds 1 \text{ or } p\) $p$ is prime


Because $H \ne K$ and $\order H = \order k$, it follows that $H \nsubseteq K$.

So:

\(\ds H \cap K\) \(\ne\) \(\ds H\) Intersection with Subset is Subset‎
\(\ds \leadsto \ \ \) \(\ds H \cap K\) \(\subset\) \(\ds H\)
\(\ds \leadsto \ \ \) \(\ds \order {H \cap K}\) \(<\) \(\ds \order H = p\)
\(\ds \leadsto \ \ \) \(\ds \order {H \cap K}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds H \cap K\) \(=\) \(\ds \set e\) Definition of Trivial Group

$\blacksquare$


Sources

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