Minimal symmetry group containing subgroup of order 2019. What is the minimal number $n$ such that symmetry group $S_n$ contains subgroup of order 2019? Is it the biggest factor of 2019, which is 673?
Edit: Sorry I missed a key property. The subgroup with order 2019 need to be a cyclic group. 
 A: Yes, it is 673. It is not a general principle that you can use the biggest prime factor though. The reasoning is specific to the fact that 2019 is a product of 2 distinct primes and the bigger one is 1 mod the smaller one. In general, the smallest such $n$ has to be at least the biggest prime factor (for the same reason as below), but it can be more. Here's the argument:
The prime factorization of $2019$ is $3\times 673$, thus the desired subgroup has $pq$-order. The problem is find the smallest $n$ such that a group of order $pq$ embeds into $S_n$.
Without loss of generality let $p<q$. (So here, $p=3$ and $q=673$.) Up to isomorphism, there are 2 groups of order $pq$ if $q=1\operatorname{mod} p$ (which holds in this case, as $672$ is a multiple of $3$): there is a cyclic group of order 2019, and the semidirect product $C_q\rtimes C_p$ where $C_p$ acts nontrivially on $C_q$. (See here for proof and examples.) The latter nonabelian group is isomorphic to the group of permutations of the points of the finite field $\mathbb{F}_q$ that have the form
$$ x\mapsto ax+b,$$
where $a$ is a $p$th root of unity in $\mathbb{F}_q$, while $b$ is any element of $\mathbb{F}_q$ at all. (A way to see why the condition $q=1\operatorname{mod}p$ is necessary for this group to exist is that this is the circumstance under which $\mathbb{F}_q$ contains the $p$th roots of unity.)
This construction realizes the nonabelian group of order $2019$ as a subgroup of $S_{673}$.
There is no smaller $n$ such that either of the groups of order $2019$ embed in $S_n$, because if $n<673$, then $S_n$'s order does not have 673 as a factor, so it cannot have a subgroup of order divisible by 673.
To see that the biggest prime factor is not enough in general, the smallest $n$ such that a group of order $p^2$ ($p$ prime) embeds in $S_n$ is $2p$. (The group generated by a pair of disjoint $p$-cycles has the right order; and no smaller $n$ allows $|S_n|$ to be divisible by $p^2$.) If $p$ and $q$ are two primes that do not satisfy $q=1\operatorname{mod}p$, then the only group of order $pq$ is the cyclic group, and this does not embed in $S_n$ for $n\leq p+q$, as there is no permutation of order $pq$ smaller than a product of a disjoint $p$-cycle and $q$-cycle.
Addendum: I see that the question has been edited to require that the embedded subgroup of order 2019 be cyclic. Dietrich Burde has answered the question in this form, and the final paragraph in this answer also implicitly answers it.
A: The smallest $n$ such that $S_n$ contains a cyclic group of
order $\prod_{p\in \Bbb P} p^{k_p}$ is given by $\sum_{p\in \Bbb P}p^{k_p}$. Hence $n=3+673$ is the smallest $n$ such that $S_n$ contains a cyclic subgroup of order $2019=3\cdot 673$. 
Any group of order $2019=pq$ is either cyclic or a semidirect product of $C_p$ and $C_q$. So we have a unique nonabelian group of order $2019$, whose minimal $n$ for an embedding to $S_n$ is given by $673$ itself, as explained in the answer above.
In general, asking about a minimal permutation representation for finite groups is difficult, and there are whole books on this subject, e.g., this one.
A: For a finite group $G$, let $\mu(G)$ be the least positive integer $n$ such that $G$ is embedded as a subgroup of the symmetric group on $n$ points. In other words, $\mu(G)$ is the minimal permutation representation degree of $G$.
The general method to find $\mu(G)$ for a finite group $G$ is this: $\mu(G)$ is the minimum of $\sum_{i=1}^t |G:H_i|$ over
all sets of subgroups $H_1, H_2, \ldots, H_t$ such that
$\bigcap_{i=1}^t\mathrm{Core}_G(H_i) = 1$, which is equivalent to
$\mathrm{Core}_G(K) = 1$, $K = \bigcap_{i=1}^tH_i$.
Figuring out what $\mu(G)$ is when $G$ is a finite abelian group of order $pq$, for distinct primes $p$, $q$, is now easy.
