Group of Prime Order p has p-1 Elements of Order p

Theorem

Let $p$ be a prime number.

Let $G$ be a group with identity $e$ whose order is $p$.

Then $G$ has $p - 1$ elements of order $p$.

Proof

Let $\order g$ denote the order of an element $g$ of $G$.

From Order of Element Divides Order of Finite Group:

$\order g \divides p$

where $\divides$ denotes divisibility.

By definition of prime number, the only divisors of $p$ are $1$ and $p$.

From Identity is Only Group Element of Order 1, only $e$ has order $1$ in $G$.

That leaves $p - 1$ elements of order $p$.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 44.2$ Some consequences of Lagrange's Theorem