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Quite possibly I'll solve this and post my answer below, but maybe others will post better answers before I get to that. Or after.$^\dagger$

The group of permuations of $\{a,b,c,d,e,f\}$ is generated by $15$ elements of order $2$, each of which is a transposition like this one: $(a,b)$. For each such transposition there are eight others with which it overlaps, as in the case of $(a,b)$ and $(b,c)$, and six others with which it does not overlap, as in the case of $(a,b)$ and $(c,d)$. Composition of two that overlap yields a $3$-cycle, and composition of two that do not overlap yields an element of order $2$.

The group of permutations of $\{a,b,c,d,e,f\}$ is also generated by $15$ elements of order $2$, each of which is a composition of three transpositions like this one: $(ab)(cd)(ef)$. For each such generator there are eight others with which does not "overlap" in a different sense than that above, such as $(ac)(be)(df)$; "ovlerlapping" in this case would mean having a $2$-cycle in common; and there are six others with which it does "overlap", as in the case of $(ab)(cd)(ef)$ and $(ab)(ce)(df)$. Composition of two that do not overlap yields a composition of two $3$-cycles; composition of two that do overlap yields a composition of two $2$-cycles.

The above is the beginning of the fact that a certain kind of bijection from the first set of $15$ generators to the second extends to an outer automorphism of the group of all $720$ permutations of $S_6$.

So I looked at certain compositions of three generators. Composing from left to right: \begin{align} (ab)(ac)(ad) & = (abcd) \\ (ab)(ac)(ae) & = (abce) \\ (ab)(ac)(af) & = (abcf) \\[10pt] (ab)(ac)(bc) & = (ac) \end{align}

  • Given the first two overlapping in $a$, there are three choices of a third one overlapping in $a$, and in each case the result is a $4$-cycle. There is only one choice of a generator overlapping with both in a different way, and the result is another generator.

So I asked myself: what fact about the second set of generators corresponds with the bulleted paragraph above? Given two generators that do not overlap, we should expect four possible choices of a third one that does not overlap either of them; in three cases the composition of the three should yield an element of order $4$ and in one case an element of order $2$ that is one of the $15$ generators. When the matter is expressed in that way, it's not hard to find them: \begin{align} \Big((ab)(cd)(ef)\Big)\Big((ac)(be)(df)\Big)\Big((ad)(bf)(ce)\Big) & = (acbe) \\ \Big((ab)(cd)(ef)\Big)\Big((ac)(be)(df)\Big)\Big((ae)(bd)(cf)\Big) & = (bfde) \\ \Big((ab)(cd)(ef)\Big)\Big((ac)(be)(df)\Big)\Big((af)(bc)(de)\Big) & = (adfc) \\[10pt] \Big((ab)(cd)(ef)\Big)\Big((ac)(be)(df)\Big)\Big((af)(bd)(ce)\Big) & = (ac)(be)(df) \end{align} So we should have another bulleted paragraph paralleling the one above:

  • Given the first two non-overlapping pairs of generators, there are three choices of a third one not overlapping with them in such a way that the result is a $4$-cycle. There is only one choice of a generator failing in a different way to overlap with both, and the result is another generator.

But (and this is my question!) what does "failing in a different way to overlap with both" mean? $$\begin{array}{cccccccc} \phantom{------------------------} \\ \hline \end{array}$$ $\dagger$ I have now posted my own answer, but maybe someone has some insights that I don't have.

Michael Hardy
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  • Typo fixed. Notice that the last $2$-cycle in the first table is the middle factor in all of the products, and in the second table the last product of three disjoint $2$-cycles is likewise the middle factor in all of the products. ${}\qquad{}$ – Michael Hardy May 30 '15 at 19:16

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In every one of the products of three generators, the first two factors are the same: $$ \Big((ab)(cd)(ef)\Big)\Big((ac)(be)(df)\Big) $$ Create a graph in which letters are adjacent to each other if they are mapped to each other by either of the two factors above. This ends up being just a circle with the following cyclic order: $$ \begin{array}{cccccc} & a & & & b \\ \\ c & & & & & e \\ \\ & d & && f \end{array} $$ The third factor must pair these off in such a way that none is paired with one adjacent to itself.

There are three isomorphic ways to do that in which there is only one pair of letters antipodal to each other. The fourth way has three pairs antipodal to each other.

Michael Hardy
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