Subgroup generated by a set of permutations in $S_6$

Question

I have a set of $22$ permutations in $S_6$. I want to know does it generates the whole $S_6$. Is there any online tool to compute it? This is the list.

 (1,2,6),
 (1,2,6)(3,4,5),
 (1,3)(2,4,6,5),
 (1,3,2,4,6,5),
 (1,3,6,5)(2,4),
 (1,3)(2,5,6,4),
 (1,3,2,5,4,6),
 (1,3,6,4)(2,5),
 (1,4,2,3)(5,6),
 (1,4,6,5)(2,3),
 (1,4)(2,3,6,5),
 (1,4,6,3)(2,5),
 (1,4)(2,5,6,3),
 (1,4,2,5)(3,6),
 (1,5,2,3)(4,6),
 (1,5,6,4)(2,3),
 (1,5)(2,3,6,4),
 (1,5,6,3)(2,4),
 (1,5)(2,4,6,3),
 (1,5,2,4)(3,6),
 (1,6,2)(3,5,4),

Answer 1

There is an online tool here. The text box at the bottom allows you to type your generating permutations, each on an individual line, like so:

(1,2,6) (1,2,6)(3,4,5) (1,3)(2,4,6,5) (1,3,2,4,6,5) (1,3,6,5)(2,4) (1,3)(2,5,6,4) (1,3,2,5,4,6) (1,3,6,4)(2,5) (1,4,2,3)(5,6) (1,4,6,5)(2,3) (1,4)(2,3,6,5) (1,4,6,3)(2,5) (1,4)(2,5,6,3) (1,4,2,5)(3,6) (1,5,2,3)(4,6) (1,5,6,4)(2,3) (1,5)(2,3,6,4) (1,5,6,3)(2,4) (1,5)(2,4,6,3) (1,5,2,4)(3,6) (1,6,2)(3,5,4)

and then press OK above the text box. The algorithm then concludes

The group $G$ is the symmetric group $S_6$ on the set $\{1,2,3,4,5,6\}$.

Answer 2

Already the three permutations $(1,2,6)$, $(1,2,6)(3,4,5)$, and $(2,5)$ in your set generate all of $S_6$. It is equivalent to show that $(1,2,6)$, $(3,4,5)$, and $(2,5)$ generate $S_6$.

We use repeatedly the fundamental fact about conjugating permutations applied to various permutations already known to be in the subgroup generated by these three elements: "The cycle decomposition of the permutation $\pi\rho\pi^{-1}$ [the conjugation of $\rho$ by $\pi$] can be obtained from that of $\rho$ by replacing each $i$ in the cycle decomposition of $\rho$ with $\pi(i)$."

For example, $(1,2,6)(2,5)(1,2,6)^{-1} = (6,5)$ and then $(1,2,6)(6,5)(1,2,6)^{-1} = (1,5)$ are both in this subgroup. Conjugating these by $(3,4,5)$, we see that every transposition $(a,b)$ where $a\in\{1,2,6\}$ and $b\in\{3,4,5\}$ is in the subgroup.

Now $(1,5)(2,5)(1,5)^{-1} = (2,1)$ and $(6,5)(2,5)(6,5)^{-1} = (2,6)$ are also in the subgroup, and hence so is $(2,1)(2,6)(2,1)^{-1} = (1,6)$; similarly, $(3,4)$ and $(3,5)$ and $(4,5)$ are in the subgroup. Since every transposition has been shown to be in the subgroup generated by $(1,2,6)$, $(3,4,5)$, and $(2,5)$, we see that they generate all of $S_6$.