I guess that there is no general solution for $C_{k}$ and unfortunately, I couldn't find a specific one for $C_{12}$.
Cayley's theorem claims that such $n$ exists, and Lagrange's theorem claims that $n \geq 4$; Don't know what to do next; If I'm not mistaken, $C_{12}$ is isomorphic to $\mathbb{Z_{12}}$ but I don't know how it may help.
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Aryaman Maithani
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Godder
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4You need an element of order $12$ in your $S_n$. There are different ways of achieving that. For instance, $S_{12}$ has a cycle of length (and therefore order) 12. Can you do better? – Arthur Jul 07 '20 at 08:34
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1Oh, well... Then $S_{7}$ may be represented as 4+3, so element order is 12 and the answer is $S_{7}$, right? – Godder Jul 07 '20 at 08:46
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You need to show that $S_6$ is impossible. But other than that, yeah. – Arthur Jul 07 '20 at 08:48
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Thanks a lot! it really helped – Godder Jul 07 '20 at 08:48
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2See http://oeis.org/A008475 – Arthur Jul 07 '20 at 08:50
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See also here. – Dietrich Burde Jul 07 '20 at 12:11
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The minimal permutation degree $d(G)$ of a finite group $G$ is the least integer $d$ such that $G$ can be embedded into $S_d$. Let $G\cong C_{p_1^{e_1}}\times \cdots \times C_{p_r^{e_r}}$. Then we have
$$ d(G)=\sum_{i=1}^r p_i^{e_i}. $$
For a proof see for example here. Since $12=2^2\cdot 3$, we have $$ d(C_{12})=2^2+3=7. $$
Dietrich Burde
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