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I'm reading Wilson's The Finite Simple Group book, mainly the Alternating group chapter, which explains the splitting criterion quite in detail. While heading there, I came across a possible permutation splitting in traspositions which Wilson suggests and I never thought of. At page 12, he depicts the (5 4 3 2) permutation, writing it in the usual 2 rows representation in which $1\rightarrow1$, $2\rightarrow5$, $3\rightarrow2$, $4\rightarrow3$ and $5\rightarrow4$. Actually he suggests to join equal numbers on top and bottom and inspecting the crossing of each string (line) between them. The first example is clear to me: actually the string joing the 5s is crossing backward all 4s, 3s and 2s strings, exactly in that order, leading the to the splitting (25)(35)(45) (priority from right). This is understood. The foggy part comes in the same page under the "The Alternating Group" chapter. In there, Wilson suggests yet another splitting in traspositions interpreting the same "string diagram" from bottom to top. Actually he suggests:

"An alternative interpretation of this picture is to read it from bottom to top, and record the positions of the strings that are swapped. In this example, we first swap the second and third strings, then the third and fourth, and finally the fourth and fifth. Thus the second string moves to the fifth position, the third string moves to the second position, and so on. In this way we have written our permutation as a product of swaps of adjacent strings."

Following its instructions, you can correctly get the following splitting: (45)(34)(23) (again from right to left).

While these product is correct and make sense to me, I cannot reconstruct it as Wilson is suggesting from the bottom to top as "swapping of strings".

I'm sure I'm missing something trivial. May you help me on that please?

Thanks in advance

Gnumbertester
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riccardoventrella
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1 Answers1

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Actually I've found a possible way to apply Wilson's words. Let's consider the bottom row of the diagram above, and label each string emanating from there from 1 to 5 (left to right). Each string crossing is swapping the two string position. So, climbing from bottom to top, we first encounter the first crossing, which swaps the 2nd and 3rd string, getting the very first transposition as Wilson claims, i.e. (2 3). After that, the core trick is consider the strings as swapped, so what you called string 2 before, you have to call it 3 now. Climbing the "new" 3rd string, we reach the very next crossing with the string 4th, getting the (3 4) transposition. With the same reasoning, the string 3 becomes the string 4, which finally crosses the last 5th string, realising the last (4 5). Building them up, you get (4 5)(3 4)(2 3) (right to left).

Notice this nails down yet another funny and quick way to split a permutation. The first one explained in my question, has a pattern like this:

(a b c d e) = (d e)(c e)(b e)(a e) (right to left)

while the new one is:

(a b c d e) = (a b)(b c)(c d)(d e) (right to left)

So, you can get a full set of splits letting the cycle roll over and replacing letters:

(2 5 4 3) = (3 4)(3 5)(2 3) (right to left)

(4 3 2 5) = (2 5)(3 5)(4 5) (right to left) ...

or:

(2 5 4 3) = (2 5)(4 5)(3 4) (right to left)

(4 3 2 5) = (3 4)(2 3)(2 5) (right to left) ...

riccardoventrella
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