Permutations of three points

Question

Permutations are being discussed in a way I haven't seen before. Not in the sense of permutations and combinations, although maybe they're the same and I'm not seeing it. The rule is that $$P_{ab}$$ swaps the elements a and b so for example $$P_{(12)}$$ means that 1 is swapped for 2 and 2 = 1. $$P_{(123)}$$ rotates elements so a becomes b becomes c which becomes a. So for example $$P_{(123)}$$ would mean replacing 1 with 2, 2 with 3 and 3 with 1. Because a = 1, b=2 and c=3, so after the operation a = 1 = 2, b = 2 = 3, c = 3 = 1, so 1=2, 2=3, 3=1. I have never seen this before and am being asked to combine two of these, so do $$P_{(12)}P_{(13)}$$ for example. Anyone seen this before and know how to do it?

I've encountered this in a course on group theory. Thanks for any help!

Answer 1

Bear in mind that the elements of the permutation group are acting on the set of symbols, so $P_{(12)} P_{(13)}$ acts on a string of symbols, say $x$. It is sometimes difficult to interpret switching numbers, so what we could do is relabel the numbers as letters of corresponding order and then switch back at the end. Of course, our set need not be either numbers or letters, merely elements of a set.

With this in mind we can see that we must work from right to left with permutations. As an example, let's see how that combination works on the string $x=(a,b,c,d,e,f,g)$:

$$ P_{(12)} P_{(13)} (a,b,c,d,e,f,g) = P_{(12)} (c,b,a,d,e,f,g) = (b,c,a,d,e,f,g).$$

Let's look at a more complicated example: $$ P_{(124)} P_{(423)} (a,b,c,d,e,f,g).$$ For this look at $P_{(423)} (a,b,c,d,e,f,g)$ first:

$$(a,b,c,d,e,f,g) \mapsto (a,d,c,b,e,f,g) \mapsto (a,c,d,b,e,f,g) \mapsto (a,c,b,d,e,f,g)$$ Then simply apply $P_{(124)}$ to the resulting string at the end: $$ P_{(124)} P_{(423)} (a,b,c,d,e,f,g) = P_{(124)} (a,c,b,d,e,f,g), $$ so $$ (a,c,b,d,e,f,g) \mapsto (c,a,b,d,e,f,g) \mapsto (c,d,b,a,e,f,g) \mapsto(a,d,b,c,e,f,g). $$