Minimal generation of $S_n$ For a given $n$, the symmetric group $S_n$ is generated by permutations $p_1$, $p_2$, and $p_3$.  Two of them generate subgroups $P_{23}$, $P_{13}$, $P_{12}$, which have group orders $a$, $b$, and $c$.
What is the minimal $max(a,b,c)$ for various $n$?  
 A: Derek Holt told me about this question yesterday evening. 
The answer is 8 for all $n > 7$, because 
(1) for every such $n$, the symmetric group $S_n$ is a smooth quotient of 
the $[4,4,4]$ Coxeter group 
$\langle \, a,b,c \ | \ a^2 = b^2 = c^2 = (ab)^4 = (bc)^4 = (ca)^4 = 1 \,\rangle$, and 
(2) for every triple $(r,s,t)$ of integers with $1 \le r \le s \le t \le 3$, 
the $[r,s,t]$ Coxeter group 
$\langle \, a,b,c \ | \ a^2 = b^2 = c^2 = (ab)^r = (bc)^s = (ca)^t = 1 \,\rangle$ 
is finite, except in the case where $(r,s,t) = (3,3,3)$, when it is 
Euclidean and therefore soluble. 
(Note: in (1) above, "smooth" means that the orders of $a, b, c, ab, bc$ and $ca$ are preserved in the quotient.) 
Fact (1) above is not difficult to prove using coset diagrams.
One construction for each of the four residue classes of $n$ mod 4 
does this for all $n > 14$, and then an easy computation in {\sc Magma}
shows the same for $n = 6, 8, 9, 10, 11, 12, 13$ and $14.$  
A: The answer is $8$ for all $n$ with $8 \le n  \le 26$ (as verret observed, it is $10$ for $n=7$), and I conjecture that it is $8$ for all larger $n$, with $S_n$ being a quotient of $$\langle x,y,z \mid x^2=y^2=z^2=(xy)^4=(yz)^4=(xz)^4=1 \rangle.$$
(It is possible that this is a known result.)
There are lots of solutions. Here are some for $n=6,7,8,9$ as requested.
$$n=6: (1, 2)(3, 4)(5, 6),\ 
    (1, 2)(3, 6),\ 
    (2, 3),$$
$$n=7: (1, 7)(2, 4)(5, 6),\ 
    (1, 2),\ 
    (1, 7)(2, 3)(4, 6),$$
$$n=8: (1, 8)(2, 3)(6, 7),\ 
    (1, 7)(2, 4)(3, 5),\ 
    (1, 2),$$
$$n=9: (1, 2)(3, 4),\ 
    (2, 4)(5, 6)(7, 8),\ 
    (3, 5)(4, 7)(6, 9).$$
