Given a finite abelian group $G$, is there a formula or quick algorithm to determine the minimum $N$ so that $G$ can be embedded into the permutation group $S_N$? If $G$ is cyclic of order $n$ I believe I know the answer: Write $n$ as a product of powers of distinct primes $p_1^{e_1}, p_2^{e_2} \ldots$ and then the minimal $N$ is the sum of these prime powers, which can be pretty small compared to $n$. Anyway, what if $G$ is simply abelian? The structure theorem for finite abelian groups should be helpful but I'm not sure how to determine the minimal $N$ for a product of cyclic groups. I can give an upper bound by representing each cyclic group as being generated by a permutation on a distinct set of characters, but I'm not sure if that gives the best possible $N$, and if it does, I don't know how to prove it.
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Write $G$ as a direct product of cyclic groups of orders $n_1,\ldots,n_k$, where each $n_i$ is a prime power. Then $G \le S_N$ with $N=n_1+n_2 + \cdots n_k$. I don't think you can do any better than that. – Derek Holt Apr 05 '14 at 19:49
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@DerekHolt Yes that's also what I alluded to at the end of my post, but I don't know how to prove that this is minimal $N$. I can prove it if $G$ is cyclic. – user2566092 Apr 05 '14 at 20:31