Write (43)(521)(51)(32)(1452)(23)(15)(125)(34) as the product of disjoint cycles. My answer is (12354), but I do not think this is correct. Will you explain?

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- Feb 7th 2013, 05:46 PMlovesmathPermutations
Write (43)(521)(51)(32)(1452)(23)(15)(125)(34) as the product of disjoint cycles. My answer is (12354), but I do not think this is correct. Will you explain?

- Feb 7th 2013, 07:09 PMjakncokeRe: Permutations
The answer i got was (1) (2345). For in this element, $\displaystyle 1 \to 2 \to 3 \to 2 \to 1 $ = $\displaystyle (1) $

- Feb 7th 2013, 07:11 PMjohngRe: Permutations
In working such a problem, you need to specify whether composition is left to right or right to left. That is, if $\displaystyle p_1$ and $\displaystyle p_2$ are permutations does $\displaystyle p_1\compose p_2$ mean first do $\displaystyle p_1$ and then $\displaystyle p_2$ or first do $\displaystyle p_2$ followed by $\displaystyle p_1$.

Example (1 2)(1 2 3)=(1 3) for left to right but (1 2)(1 2 3)=(2 3) for right to left. I think probably most algebraists do composition left to right.

Assuming left to right, I get (1 3 2 4)(5)

Good case in point. The previous answer does composition right to left! - Feb 7th 2013, 07:16 PMjakncokeRe: Permutations
I've always assumed the standard used in university was to do it from right to left but john makes a very good point to keep in mind.

- Feb 7th 2013, 07:47 PMlovesmathRe: Permutations
We are doing them from right to left.

- Feb 8th 2013, 12:31 AMDevenoRe: Permutations
it is a well-known fact that if τ is a cycle: say τ = (a

_{1}a_{2}a_{3}....a_{k}),

then στσ^{-1}= (σ(a_{1}) σ(a_{2}) σ(a_{3})....σ(a_{k}))

for example, if τ = (1 3 4 2), and σ = (1 2 3), so that:

σ(1) = 2

σ(2) = 3

σ(3) = 1

σ(4) = 4

then στσ^{-1}= (σ(1) σ(3) σ(4) σ(2)) = (2 1 4 3) (this assumes we are working "right-to-left" so στ(a) = σ(τ(a)), as with functional composition).

in the permutation to be evaluated at hand we have the 4-cycle:

(1 4 5 2) which is conjugated by the transposition (2 3), that is:

(3 2)(1 4 5 2)(2 3) = (2 3)(1 4 5 2)(2 3) = (2 3)(1 4 5 2)(2 3)^{-1}(transpositions are of order 2, and thus are their own inverses).

the net effect of this is to replace the "2" in (1 4 5 2) with a 3:

(2 3)(1 4 5 2)(2 3)^{-1}= (1 4 5 3).

we then have this conjugated by (1 5) = (5 1), which has the effect of making the 1 and 5 switch places, giving:

(5 1)(1 4 5 3)(1 5) = (1 5)(1 4 5 3)(1 5) = (1 5)(1 4 5 3)(1 5)^{-1}= (5 4 1 3) = (1 3 5 4).

we then have THIS conjugated by (1 5 2) (note that (1 5 2) = (5 2 1)). so in (1 3 5 4) we replace 1 by the image of 1 under (1 5 2), which is 5; the 3 stays, since 3 is not moved by (1 5 2), the 5 is replaced by 2 (2 is the image of 5 under (1 5 2)), and finally the 4 is unaffected:

(5 2 1)(1 3 5 4)(1 2 5) = (1 5 2)(1 3 5 4)(1 2 5) = (1 5 2)(1 3 5 4)(1 5 2)^{-1}= (5 3 2 4) = (2 4 5 3).

again, this is conjugated by the transposition (3 4) = (4 3), so the 2 and 5 in (2 4 5 3) are unaffected, and the 3 and 4 change places, giving:

(2 3 5 4) (it is customary, but not required, to put the least element of a cycle first). there is no reason to include 1-cycles in a disjoint cycle decomposition, so they are typically omitted.

one can also do this "the long way" calculating each successive permutation in turn (starting with (3 4) as the first permutation applied, followed by (1 2 5) and so forth):

1-->1-->2-->2-->3-->3-->2-->2-->1-->1

2-->2-->5-->1-->1-->4-->4-->4-->4-->3

3-->4-->4-->4-->4-->5-->5-->1-->5-->5

4-->3-->3-->3-->2-->1-->1-->5-->2-->2

5-->5-->1-->5-->5-->2-->3-->3-->3-->4

which also gives the 4-cycle (2 3 5 4).