The answer i got was (1) (2345). For in this element, =
In working such a problem, you need to specify whether composition is left to right or right to left. That is, if and are permutations does mean first do and then or first do followed by .
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!
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).