# Understanding an Outer Automorphism of $S_6$

In an article (paper), there is a description of an outer automorphism of $S_6$. There are six pentagons, arranged with a rule, with vertices $1,2,3,4,5$. Any permutation of these vertices will permute the six pentagons, hence giving a map (homomorphism) from $S_5$ into $S_6$.

I understood this map as follows: if we interchange the vertices, the pentagons will be permuted. Consider permutation $(2 \,3)$, and its effect on the first pentagone a in the note.

(1) In a, vertices $2,3$ are not joined by a "continuous edge", hence after permuting $2,3$ there should not be continuous edge between $2,3$. Hence image of a after this permutation will be either a,d or f (am I correct?). Also, 5 is joined to both $2,3$ before permutation, hence after permuting $2,3$, vertex $5$ should be adjcent to them. We can conclude that a is mapped to a by permutation $(2\,3)$.

(2) The other way, in a, $2,3$ are not joined by continuous edge before permutation $(2\,3)$. Hence after permuting $2,3$, a would be mapped into either a,d or f. Also, $4$ is joined to $2$ but not $3$ before permuting $2,3$; after permuting $2,3$, the vertex $4$ will be joined to $3$ but not $2$; hence image of a by permutation $(2\,3)$ should be d.

I couldn't find my mistake in understanding, if any.

Can one explain the action of $S_5$ on the six pentagons in the article?

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You've omitted an essential fact without which this is incomprehensible. The point is that there are six ways to 2-color the vertices. Without that, I don't see where your six pentagons come from. – Michael Hardy Aug 9 '12 at 18:47

The crucial part you missed is "(up to choices of colors)" (in the third line of the section). If you distinguish between colours, there are twice as many pentagons. Applying $(23)$ to a yields e with colours swapped.
@jorki: OK; means, given a (ordered) 5-cycle on vertices $\{1,2,3,4,5\}$, there is unique pentagon containing the 5-cycle. By permutation of vertices, these 5 cycles (and hence the pentagons) will be permuted; this will be the map (homomorphism) from $S_5$ to $S_6$; am I correct? – Groups Aug 10 '12 at 4:54