The order of an element and its inverse are the same

Given a group $𝐺$ and an element $𝑎 \in 𝐺$ with order $| 𝑎 | = 𝑛$ , the order of the inverse $∣ 𝑎 - 1 ∣ = 𝑛$ . #m/thm/group

Proof

By the uniqueness of the inverse, $𝑎 𝑛 = 𝑒$ iff $𝑎 - 𝑛 = 𝑒$ . Therefore the orders of $𝑎$ and $𝑎 - 1$ must be equal, since neither can have a lower order than the other.

#state/tidy | #lang/en | #SemBr