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For every positive integer $m$, consider the symmetric group $S_m$. By a full cycle, I mean a permutation $\sigma\in S_m$ which is a single cycle of length $m$. I will say two permutations $\sigma$ and $\tau$ are orthogonal if $\sigma^{-1}\tau$ has no fixed points, or in other words, $\sigma(x)\neq\tau(x)$ for all $x\in\{1,2,\ldots,m\}$. I want to know, for each $m$, what the maximum number of pairwise orthogonal full cycles in $S_m$ is. That is, what is the maximum number $n$ for which there exist $\sigma_1,\sigma_2,\ldots,\sigma_n\in S_m$ such that each $\sigma_i$ is an $m$-cycle and for any distinct $i$ and $j$, $\sigma_i^{-1}\circ\sigma_j$ is fixed-point free?

It is easy to show that $n\leq m-1$ for $m\geq 2$. This is simply because for any $x\in\{1,2,\ldots,m\}$, the assumptions imply $\sigma_1(x),\sigma_2(x),\ldots,\sigma_n(x)$ are distinct and $\{\sigma_1(x),\sigma_2(x),\ldots,\sigma_n(x)\}\subseteq\{1,2,\ldots,m\}\setminus\{x\}$. We can also show that $n=m-1$ if $m$ is prime since then we can take $\sigma_i$ to be the function which adds $i$ modulo $m$, which are all $m$-cycles because $1,2,\ldots,m-1$ are all coprime to $m$ when $m$ is prime. I believe the converse is true too: If $m$ is a composite number, then $n\leq m-2$.

In all of the examples I have checked, $n=m-2$ when $m$ is composite. So I really just want to know if this is true for all composite numbers.

Anon
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    For a lower bound we may essentially use the Chinese remainder theorem. For any composite $m=p_1^{\alpha_1}\cdot\ldots\cdot p_k^{\alpha_k}$ and any $\eta\in[1,k]$ we may partition $(1,\ldots,m)$ into $p_\eta^{\alpha_\eta}$ blocks and define a map $\psi_{\eta,i}$ which brings a block with index $a$ into a block with index $a+i\pmod{p_\eta^{\alpha_\eta}}$. If $i\neq 0$ this map has order $p_{\eta}^{\alpha_{\eta}}$, hence by composing different $\psi_{\eta,i}$ for $\eta$ ranging from $1$ to $k$ we get at least $\prod_{\eta=1}^{k}\left(p_\eta^{\alpha_1}-1\right)$ orthogonal cycles. – Jack D'Aurizio Aug 12 '25 at 17:27
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    Maybe by combining shifts and blocks-shifts it is also possible to reach $m-2$: too look for a suitable "tensor trick" seems the way to go. – Jack D'Aurizio Aug 12 '25 at 17:54
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    @JackD'Aurizio There is one minor error with your calculation. In order for $\psi_{\eta,i}$ to have order $p_{\eta}^{\alpha_{\eta}}$, you need $i$ to be coprime to $p$, not simply $i\neq 0$. Thus, you actually get $\prod_{\eta=1}^k(p_{\eta}^{\alpha_{\eta}}-p_{\eta}^{\alpha_{\eta}-1})$. Note however, that this is just $\phi(n)$ where $\phi$ is the Euler totient function. We could achieve this more simply by letting $\sigma_i$ be addition by $i$ modulo $m$ where $i$ is coprime to $m$. – Anon Aug 12 '25 at 18:14
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    you're right. Still we may consider a standard shift for any $(a,m)=1$ and maybe a suitable composition of block-shifts if $(a,m)>1$. – Jack D'Aurizio Aug 12 '25 at 20:09

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We can naturally interpret your question as on a maximal number of pairwsie edge disjoint Hamiltonian cycles in a complete directed graph with $m$ vertices. As I understand this answer, this number is $m-1$ for each natural $m>1$, but $4$ and $6$.

Alex Ravsky
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