The Cayley-group-theorem states that every group is isomorphic to a subgroup of a permutation group. I am especially interested in the case that the group is finite.
My question :
If G is a group with order $n$ , what is the smallest number $k$, such that $G$ is isomorphic to a subgroup of the permutation group $S_k$ ? Do we need $k=n$ in general, or is it always sufficient to take the smallest number $k$ with $n|k!$ ?
$S_m$ has $m!$ elements but is a subgroup of $S_k$ with $k=m$. So, you don't always need $k=|G|$.
On the other hand, let $p$ be prime greater than 2 (if $p=2$, then $p=p!$). Observe that $\mathbb{Z}/p\mathbb{Z}$ is cyclic of order $p$. However, the smallest $n$ such that $S_n$ has a cyclic subgroup of order $p$ is $n=p$ (because you need a $p$-cycle to get an element of order $p$).
Edit: To address the question in your comment: yes, every finite group $G$ is a subgroup of $S_k$, where $k=|G|$.
$D_8$ the dihedral group of order $8$ is a subgroup of $S_8$ by Cayley theorem. But $D_8$ is also a subgroup of $S_4$. This example shows that Cayley theorem is not the best bound.
You also wonder that is it enough $n | k! $ ? Answer is no. For example $6$ divides $4!$ but $Z_6$ is not a subgroup of $S_4$.
In genereal it is not known that smallest $k$ such that $G\leq S_k$.
But there are some useful lemmas, in some cases;
Lemma1: Let $G$ be simple group and $H$ be a subgroup of index $n$. Then $G\leq A_n$.
Lemma2: Let $G$ be group and $H$ be a subgroup of index $n$ such that $H$ does not contain any nontrivial normal subgroup of $G$. Then $G\leq S_n$.
This paper is mainly related to your problem; link.
I am not sure if there is a broader theory that address this question, but we certainly do not need $k = n$ in general. For example, $\mathbb{Z}/6\mathbb{Z}$ is isomorphic to the subgroup $\left<(12)(345)\right>$ of $S_5$. However, there also are cases in which you must take $k \geq n$. For example, it can be shown that $Q_8$ is not isomorphic to any subgroup of $S_k$ for any $k < 8$.
The other answers show that $k=\vert G\vert$ can be necessary; let me fill in the argument that it is always sufficient.
To see that $G$ is always (isomorphic to) a subgroup of $S_n$ - where $n=\vert G\vert$ - we can argue as follows:
First, list $G$ as $\{g_1, g_2, . . . , g_n\}$.
Now, each element $g\in G$ induces a permutation $\pi_g$ of $\{1, . . . , n\}$ as follows: $$\pi_g(i)=j\iff g\cdot g_i=g_j,$$ where $\cdot$ is the group operation.
This gives an injective map from $G$ to $S_n$: $g\mapsto \pi_g$. It remains to show that this map is a homomorphism; this is a good exercise.
