Intersection of Subgroups of Prime Order

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

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 43$. Lagrange's theorem: Worked Example $1$