# Group has Subgroup of Prime Order if Prime Divides Order

 It has been suggested that this page or section be merged into Cauchy's Group Theorem. (Discuss)

## Theorem

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

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

Then $G$ has an element whose order is $p$.

Consequently, $G$ 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 $\phi: G \to G'$.

Then:

$\left({h'}\right)^p = e'$

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

Thus $h^p \in \gen g$ so $\left({h^p}\right)^n = \left({h^n}\right)^p = e$.

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

If $h^n = e$ then $\left({h'}\right)^n = e'$.

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

$\blacksquare$

## Proof 2

This result obtains as a special case of Subgroups of All Prime Power Factors.

$\blacksquare$