Cauchy's Group Theorem

From ProofWiki
Jump to navigation Jump to search

This proof is about Cauchy's Group Theorem. For other uses, see Cauchy's Theorem.


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

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

Then $G$ has a subgroup of order $p$.

Proof 1

Let $\order G$ be a prime number.

Then from Prime Group is Cyclic $G$ has a generator $\gen g$ such that $\order g = p$.

Now suppose $e \ne g \in G$.

Let $\order g = n$.

Let $p \divides n$.

Then by Subgroup of Finite Cyclic Group is Determined by Order the cyclic group $\gen g$ has an element $g$ of order $p$.

Suppose $p \nmid n$.

From Subgroup of Abelian Group is Normal, $\gen g$ is normal in $G$.

Consider the quotient group $G' = \dfrac G {\gen g}$.

As $p \nmid n$, $\gen e \subsetneq \gen g \subsetneq G$.

Thus $\order {G'} < \order G$.

But we have that $p \divides \order G$.

It follows by induction that $G'$ has an element $h'$ of order $p$.

Let $h$ be a preimage of $h'$ under the quotient epimorphism $\phi: G \to G'$.


$\paren {h'}^p = e'$

where $e'$ is the identity of $G'$.

Thus $h^p \in \gen g$ so $\paren {h^p}^n = \paren {h^n}^p = e$.

Thus either $h^n$ has order $p$ or $h^n = e$.

If $h^n = e$ then $\paren {h'}^n = e'$.

But since $p$ is the order of $h'$ it would follow that $p \divides n$, contrary to assumption.


Proof 2

This result follows as a special case of Group has Subgroups of All Prime Power Factors.


Also known as

Cauchy's Group Theorem is also seen referred to as Cauchy's Theorem, but as there are several theorems named so, this is not endorsed by $\mathsf{Pr} \infty \mathsf{fWiki}$.

Source of Name

This entry was named for Augustin Louis Cauchy.

Historical Note

Cauchy's Group Theorem was proved by Augustin Louis Cauchy in $1845$.