Cauchy's Lemma (Group Theory)/Proof 2

Theorem

Let $\struct {G, \circ}$ be a group of finite order whose identity is $e$.

Let $p$ be a prime number which divides the order of $G$.

Then $\struct {G, \circ}$ has an element of order $p$.

Proof

By the corollary to the First Sylow Theorem, $G$ has subgroups of order $p^r$ for all $r$ such that $p^r \divides \order G$.

Thus $G$ has at least one subgroup $H$ of order $p$.

As a Prime Group is Cyclic, $H$ is a cyclic group.

Thus by definition $H$ has an element of order $p$.

Hence the result.

$\blacksquare$

Source of Name

This entry was named for Augustin Louis Cauchy.

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.19$
• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 56$. First Sylow Theorem