TITLE: Finding the minimal $n$ such that a given finite group $G$ is a subgroup of $S_n$
QUESTION [15 upvotes]: It is a theorem that, for every finite group $G$, there exists an $n$ such that $G \leq  S_n$. Given a particular group, is there any way to determine or bound the smallest $n$ such that this occurs? The representation of the group is up to you, though bounds that use less information are preferred.
If this problem is too hard, specific special cases are also interesting, as are bounds on $n$.

REPLY [2 votes]: Given a finite group $G$, consider $H$ to be subgroup such that it contains no normal subgroup of $G$ and consider such $H$ of maximum possible order. 
Then $G$ permutes cosets of $H$ by left multiplication, and hence gives a homomorphism from $G$ to permutation group of cosets $\{gH\colon g\in G\}$. The kernel of this homomorphism is intersection of all the conjugates of $H$; but, by our assumption, it should be trivial. Hence $G$ embeds in permutation group on $|G/H|$ letters. 
This also explains why $H$ is chosen in beginning of maximum order.
This works, as it can be seen for examples in other answers.