Clearly the identity function $f =\text{id}$ satisfies $f^n = \text{id}$ for any $n \in \mathbb{N}$. However, there are also other functions with this property. For instance, with $n=2$ we have self-inverse functions like $x \mapsto 1-x$. For functions $f:\mathbb{R}\rightarrow\mathbb{R}$, we can use roots of unity e.g. $f(x) = \exp({\frac{2\pi i}{n}})\times x$.

I am interested in functions $f:\mathbb{N} \rightarrow \mathbb{N}$, and in particular $n=3$. Are there non-identity functions such that $f \;\circ f \;\circ f =\text{id}$? If so, how many functions of this type? Can you give an example, or an infinite family of examples?

I am not looking for rigorous proofs here, but just examples. Probably there is a name for functions of this type but I have searched and cannot find anything. This is not part of any formal teaching, but just an interesting question I came up with.

  • 1
    $\begingroup$ You could consider permutations, since these are precisely bijections from a subset $\Omega$ of $\mathbb{N}$, and defining a permutation of a particular order in this case is easy. They can be extended to all of $\mathbb{N}$ by sending each value in $\mathbb{N}\backslash\Omega$ to itself, or by adding more $n$-cycles to the permutation if you want a permutation of order $n$. $\endgroup$ – Bill Wallis Apr 30 '18 at 13:39

Any partition of $\mathbb{N}$ into ordered triples gives a function $f$ that acts on each triple by cycling $(x, y, z) \mapsto (y, z, x)$ (and conversely, every function $f$ can be described fully by such a list of triples) This gives you an uncountably infinite family of functions. If you want an explicit function of this type, consider $$f(n) = \begin{cases} n - 2 & \text{$n$ is a multiple of 3} \\ n+1 & \text{otherwise} \end{cases} $$


Such things are called functional $n^{th}$ roots (of the identity) which perhaps you already knew. I believe it's possible to show there are infinitely many such things and of fairly bizarre nature using Zorn's lemma (actually I guess you don't even need this).

One example is

$f(1) = 2, f(2) = 3, f(3) = 1$,

$f(4) = 5, f(5) = 6, f(6) = 4$



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.