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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
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1 Answer

up vote 4 down vote accepted

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.

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@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
    
@user28017: Yes, correct! –  joriki Aug 10 '12 at 8:05
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