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Given the symmetric group $S_n$, consider a chain of subgroups

$$S_n=G_k>G_{k-1}>\cdots>G_2>G_1>G_0=1$$

and let

$$m=\max_{1\leq j\leq k}\frac{|G_j|}{|G_{j-1}|}.$$

Define $\chi(S_n)$ as the minimum value of $m$ over all such chains.

Clearly $\chi(S_n)\leq n$, as shown by the chain

$$S_n>S_{n-1}>\cdots>S_2>S_1=1.$$

And $\chi(S_n)\geq p$, for any prime $p$ dividing $|S_n|=n!$, that is, any prime $p\leq n$. It follows that $\chi(S_p)=p$, for any prime $p$.

For $S_4$, thinking geometrically, I recalled that the regular tetrahedron symmetry group contains the digonal antiprism symmetry group, of order $8$ (index $3$), which in turn contains a group of order $4$ and then $2$ (indices $2\leq3$). Hence $\chi(S_4)=3$.

Do we always have $\chi(S_n)=\max\{\text{primes }p\leq n\}$? Or is there some other general formula? If not, at least I want the values of $\chi(S_n)$ for $n\leq16$.

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1 Answer 1

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For any $n \geqslant 5$, we know that $A_n$ is simple. You can use this to show that any proper subgroup $H$ of $A_n$ has index $k \geqslant n$. Indeed, we have an action of $A_n$ on the left cosets of $H$, giving a map $A_n \to S_k$ which is non-trivial (since $A_n$ acts transitively on the cosets), and since $A_n$ is simple, this must be an injection, giving $\tfrac{n!}{2} \leqslant k!$, so $k \geqslant n$. By a similar argument, you can show that the only subgroup of $S_n$ of index $k < n$ is equal to $A_n$ (can you see why?) This will prove that $\chi(S_n) = n$ for all $n \geqslant 5$.

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