Is $\langle(26543),(34)(56),(12)(3654)\rangle $ isomorphic to $A_6$?

Question

My question is the one in the title.

Is $\langle(26543),(34)(56),(12)(3654)\rangle $ isomorphic to $A_6$?

Answer 1

The subgroup is $A_6$, which we can show by proving that every product of two disjoint transpositions is in the subgroup (and observing that the three generators are all even). More precisely, every product of two disjoint transpositions is a conjugate of $(34)(56)$.

I propose visualizing the group as permuting the points 1 through 6 arranged as this:

          6-----5
        / :     |
  1----2  :     |
        \ :     |
          3-----4

The generator (26543) cyclically rotates the pentagon that consists of all points except 1. The generator (12)(3654) cyclically rotates the square while swapping 1 and 2.

Now in order for a product of two disjoint transpositions to be a conjugate of (34)(56), what we must show is that if we place two red markers on some of the positions and two green markers on some other positions, it is always possible to apply some of the generator permutations such that the markers of one color end up at 5 and 6, and the other color at 3 and 4:

          G-----G               R-----R
        / :     |             / :     |
  x----x  :     |  or   x----x  :     |
        \ :     |             \ :     |
          R-----R               G-----G

This can be done in steps:

First, if there's a marker at 1, apply (26543) until there's no marker at 2, and then apply (12)(3654) once.

Now all four markers are somewhere among {2,3,4,5,6}, and one or more applications of (12)(3654) and/or (26543) will lead to one of these states (or their counterparts with red and green swapped):

          R-----R           G-----R
        / :     |         / :     |
  x----x  :     |   x----R  :     |
        \ :     |         \ :     |
          G-----G           G-----x

In the left case, we're done. In the right one (34)(56) followed by (26543) leads to the left one.

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Alternatively, we can first show that (34)(56) and (26543) generate all even permutations of {2,3,4,5,6}. It is enough to show that they generate all products of two transpositions. If the two transpositions are disjoint, then their product is $(26543)^k(34)(56)(26543)^{-k}$ for some $k$. On the other hand, if they are not, then rewrite as $(ab)(bc)=(ab)(ef)(ef)(bc)$ (where $e$ and $f$ are the two remaining elements), and then express $(ab)(ef)$ and $(ef)(bc)$ separately.

Then every permutation from $A_6$ can be created by first using some power of $(26543)$ followed by $(12)(3654)$ to put the desired element at position 1; then the remaining rearrangement is an even permutation of {2,3,4,5,6}, which we have just shown we can do.