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Let $G$ be a finite abelian group. Write $Gapprox mathbbZ/p_1^i_1mathbbZtimesdots mathbbZ/p_m^i_mmathbbZ$, with $mge 0$, $p_1,dots,p_m$ primes (not necessarily distinct) and $i_kge 1$ for all $k$.

For $nge 0$, is it true that there is an injective homomorphism $Grightarrow S_n$ if and only if $p_1^i_1+dots+p_m^i_mleq n$?

Yes, this is true (provided we add the hypothesis $i_1, ldots, i_n ge 1$), and not hard to prove by considering centralizers of products of cycles in symmetric groups.

According to http://www-history.mcs.st-andrews.ac.uk/Biographies/Povzner.html, the minimum degree of a permutation representation of an arbitrary abelian group was found in A. Ya. Povzner, Finding groups of permutations of least degree which are isomorphic to a
given Abelian group, Kharkov Mat. Obshch., 14 (1937) 151–158. (I have not been able to obtain this paper to check.)

Theorem 4.1 in Kay Jin Lim, Specht modules with abelian vertices,
J. Algebraic Combin. 35 (2012) 157–171 states your conjectured result in the special case when all the primes are the same with a reference to Povzner's paper.

Yes. Your notation $n$ is used for two different meanings, so let me denote by $m$ the number of summands. A homomorphism $Gto S_n$ correspond to an action of $G$ on a set with $n$ elements, which can be split into orbits $A_1,...,A_ell$. Each orbit can can be provided with a structure of a group with a surjective homomophism $Gto A_i$. Then, the condition is precisely that the induced map $Gto oplus_i A_i$ is injective. Note the total number of cyclic summands in the right hand side is at least $n$. If $A_i$ is not cyclic, using $abge a+b$ for $a,b>1$, we see that it only decreases the sum of sizes if we split $A_i$ into two parts corresponding to a decomposition $A_i = Boplus C$. So we may assume that all the $A_i$-s are cyclic of order a prime power. By comparing each primary component at a time, we may assume that all the $p_i$-s are a single prime $p$. The result now follows from the following standard lemma

Lemma: Let $phi: oplus_i mathbbZ/p^a_i to oplus_j mathbbZ/p^b_j$ be injective, with $a_1 ge a_2 ge ...ge a_k$ and $b_1 ge ...ge b_ell$. Then $a_i le b_i$.

edit: As mensioned correctly by @Richard Lyons, the proof I gave for it was totally incorrect so I deleted it. Here's an hopefully less wrong proof, im really sorry for that.

Let $Asubseteq B$ be abelian finite $p$-groups. Write $B=mathbbZ/p^k oplus B'$ for $B'$ of exponent smaller than or equal $k$. Let $Bto mathbbZ/p^k$ be the projection. If the composition
$Ato Bto mathbbZ/p^k$ is not surjective, then $A$ is contained in $pmathbbZ/p^k oplus B$ and we can proceed by induction onthe total number of elements of $B$. Otherwise $A$ contains an element of the form $(1,b)$ for $bin B'$. Since the exponent of $B'$ is no more than $k$, there is a map $mathbbZ/p^kto B'$ mapping $1$ to $b$, and so by applying the automorphism $(x,c)mapsto (x,c-xb)$ we may assume that $(1,0)$ is in $A$. It follows then that $A=mathbbZ/p^k oplus Acap B'$ and we proceed by induction on the number of summands.