Simple 4-cycle permutation

I call a 4-cycle permutation simple if I can write it as $(i,i+1,i+2,i+3)$ so $(2,3,4,5)$ is a simple 4-cycle but $(1,3,4,5)$ is not. I want to write $(1,2,3,5)$ as a product of simple 4-cycles. So this is what I did: $$(1,2,3,5)=(1,2)(1,3)(1,5)$$ but \begin{align} (1,3)&=(2,3)(1,2)(2,3)\\ (1,5)&=(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5) \end{align} So now $$(1,2,3,5)=(1,2)(2,3)(1,2)(2,3)(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5)$$ Can you please give me a hint on how I can express $$(1,2)(2,3)(1,2)(2,3)(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5)$$ as a product of simple 4-cycles.

Note: We do permutation multiplication from left to right.

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One way would be to write $(1,2,3,5) = (4,5)(1,2,3,4)(4,5)$ and try to write $(4,5)$ as a simple $4$-cycle instead of trying to do so for all the 2-cycles you came up with.

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That is my main problem. I do not know how could I write $(4,5)$ as a product of simple 4-cycles. –  Vafa Khalighi Mar 22 '11 at 3:26

I'm strictly shooting from the hip (i.e. this is just instinct), but it might help if you consider the following:
(1) The 4th powers of 4-cycles are unity. I.e. $(1234)^4 = e$ where $e$ is the identity.
(2) This means that inverses exist. I.e. $(1234)^3 (1234)^1 = e$.
(3) And this gives a suggestion for a way of walking around the 4-cycles.

If you need another hint, come back and ask again tomorrow?

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But $(1,2,3,4)^{-1}=(4,3,2,1)$ is no longer a simple 4-cycle. –  Vafa Khalighi Mar 22 '11 at 3:25
@Vafa: By Carl's comment, $(4,3,2,1) = (1,2,3,4)^3$. –  Alex Becker Mar 22 '11 at 3:30