What is the name of an operator $T$ satisfying $T^n=T$?

Question

What is the name of an operator $T$ satisfying $T^n=T$? If $n=2$, we say that $T$ is idempotent. But what about for $n>2$?

Answer 1

Your operator is cyclic of order $n$. (There are three cases: (1) $T$ can be cyclic of finite order, (2) $T$ can be cyclic of infinite order, meaning that it is invertible but with $T^n \neq T^{-1}$ for any $n \in \Bbb{N}$ or (3) $T$ is not invertible, e.g., $T : \Bbb{Z} \to \Bbb{Z}$ with $T(i) =|i|$.)

Answer 2

I don't know of any special name for such operators, but they do all satisfy one engaging property related to idempotence: if

$T^n = T, \; n \ge 2, \tag 1$

then

$(T^{n - 1})^2 = T^{ n - 1}, \tag 2$

that is, $T^{n - 1}$ is idempotent, seen as follows:.

$(T^{n - 1})^2 = T^{2n - 2} = T^n T^{n - 2} = T T^{n - 2} = T^{n - 1}; \tag 3$

so even though $T$ is not itself idempotent, it is closely related to one.

I guess you could call such $T$ an "$(n - 1)$-st root of an idempotent".