# Number of Order p Elements in Group with m Order p Subgroups

## Theorem

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

Let $G$ have $m$ subgroups of order $p$.

The total number of elements of $G$ of order $p$ is $m \paren {p - 1}$.

## Proof

Let $H \le G$ be a subgroup of $G$ of order $p$.

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

From Group of Prime Order p has p-1 Elements of Order p, $H$ has $p - 1$ elements of order $p$.

From Intersection of Subgroups of Prime Order, each of the $m$ sets of $p - 1$ non-identity elements of the $m$ subgroups of order $p$ are pairwise disjoint.

Each elements of $G$ order $p$ is the generator of a cyclic group of order $p$.

Therefore there are no elements of order $p$ apart from the $m \paren {p - 1}$ such elements of the $m$ subgroups of order $p$.

Hence the result.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $19$