# What are the best known asymptotic bounds on the size of the largest non-trivial subgroup of the symmetric group?

To rephrase, how big can $$\frac{|H|}{n!}$$ get when $$H$$ is a non-trivial subgroup of $$S_n$$?

A trivial upper bound is $$\frac{1}{2}$$ due to Lagrange's theorem. An obvious, lazy lower bound is $$\frac{1}{n}$$ since $$S_{n-1}$$ is isomorphic to a subgroup of $$S_n$$. What better asymptotic bounds are known/can be shown?

Edit: As pointed out in the comments by Chris Leary, the alternating subgroup $$A_n$$ achieves $$\frac{1}{2}$$, which I somehow completely missed.

• The alternating group $A_n$ of even permutations is an index $2$ subgroup of $S_n$. It actually achieves your upper bound of $1/2$. You must decide if there is a nontrivial subgroup of greater cardinality than $n!/2$. Commented Jul 20 at 18:30

Since the alternating subgroup trivializes this question for $$S_n$$ we can re-ask the question for $$A_n$$. We always have a subgroup $$A_{n-1}$$ of index $$n$$ (for $$n \ge 3$$). If $$G$$ is a subgroup of index $$d$$, then $$A_n$$ acts transitively on the cosets $$A_n/G$$ and hence admits a nontrivial homomorphism
$$A_n \to \text{Aut}(A_n/G) \cong S_d$$
which for $$n \ge 5$$ must be injective since $$A_n$$ is simple. It follows that $$d \ge n$$. So it is not possible to do better than index $$n$$ for $$n \ge 5$$.