The order of an element divides the order of a finite group. This divisibility has an important power consequence. If $a$ is an element of a finite group $G$, then raising $a$ to the power $|G|$ gives the identity element. This result comes directly from the cyclic subgroup generated by $a$. In this lecture, we prove the formula, compute examples, and clarify how the result depends on the finite nature of the group.

The equation $a^{|G|}=e$ comes from the smaller cycle controlled by $o(a)$. Choose an element order and a multiple of that order to make a possible group order. The cycle marks powers by their exponent modulo $o(a)$, so every multiple of $o(a)$ returns to the identity. Notice that $|G|$ works because it is a multiple of $o(a)$, not necessarily because it is the smallest exponent that returns to $e$.

The visual cycle separates the element order from the group order. The group order may travel around the same cycle several times before landing back at the identity.

The formula $a^{|G|}=e$ should not be confused with the statement that $|G|$ is the order of every element. The actual order $o(a)$ may be a proper divisor of $|G|$. For example, in a group of order $12$, an element may have order $1$, $2$, $3$, $4$, $6$, or $12$, depending on the group and the element.

Powers of an element repeat with period $o(a)$. Enter an element order and an exponent to reduce the exponent modulo the order. If the remainder is $0$, the power is the identity element. Otherwise, the calculator gives the equivalent reduced power.