# Roots of Identity Matrix

I was reading through some problems based on permutations. Particularly, the following problem interested me.

Let $$\mathbb{Z}_n$$ denote the set of integers $$\{1,2,\ldots,n\}$$ for any positive integer $$n$$ and let $$f : \mathbb{Z}_n \rightarrow \mathbb{Z}_n$$ be a bijective function defined on $$\mathbb{Z}_n$$.

Given that $$n$$ is finite, does integer $$k^*$$ exist, such that it is always possible to find a $$k \le k^*$$ for which $$f^k(x)=x, \forall x \in \mathbb{Z}_n$$?

I suspect that this would turn out to finding the primitive real roots of the identity matrix. That is finding the maximum value of $$k$$ such that $$X^k=I_n$$ has a $$n \times n$$ matrix solution $$X$$ , but, $$X^i \neq I_n$$ for any $$i < k$$

Taking into account that any arbitrary permutations could consist of $$m$$ cycles of length $$i_1,i_2,\cdots,i_m$$ with $$i_1+i_2+\cdots+i_m=n$$. The problem now boils down to finding the maximum value of the LCM of $$i_1,i_2,\cdots,i_m$$ over all possible values of $$m$$ i..e,

$k^* = \max_{\substack{m=1\cdots n\\ i_1+i_2+\cdots+i_m=n}} LCM (i_1,i_2,\cdots,i_m)$ But, am at a loss in actually proceeding much further.

Note by Janardhanan Sivaramakrishnan
1 year, 11 months ago

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