# 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.

## 3 Answers

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.

• 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

(2,3,4,5) is (1,2,3,5) conjugated by (1,2,3,4) so (1,2,3,5) = (1,2,3,4)(2,3,4,5)(1,2,3,4)−1.

While this may appear to be coincidence, I think you'll find it works quite well in general as suggested by Carl Brannen.

• 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
• Thanks. I really hate myself. I am too dumb! – Vafa Khalighi Mar 22 '11 at 3:31
• @Alex: One more question. How do you know that "(2,3,4,5) is (1,2,3,5) conjugated by (1,2,3,4)". When I look at this, it makes sense, but say you only have (1,2,3,5), how do you find out that "(2,3,4,5) is (1,2,3,5) conjugated by (1,2,3,4)". – Vafa Khalighi Mar 22 '11 at 4:02
• Conjugating by a permutation is the same as applying that permutation inside each cycle, and then it is often easy to see how to get one permutation from another by conjugating. – Tobias Kildetoft Mar 22 '11 at 10:10

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?