# Multiplying Cycle Permuations

I am having trouble multiplying permutations in cycle notation.

(1 3 4 5) (2 3 4) = (1 3 5) (2 4)

I do not understand how this product is determined. My answer is (1 3 4) (2 5). I have come to this conclusion by switching the order to (2 3 4) (1 3 4 5) (because it is a composition) then: begin with 1. 1--> 3 next 2 --> 3 and 3 --> 4. Then 2 --> 4 and 4 --> 5.

Can anyone explain the flaw in my method?

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$(1 3 4 5)(2 3 4) \neq (2 3 4)(1 3 4 5)$ – user96614 Oct 20 '13 at 5:26
Ok, but in that case that case the solution i get is (1 4 3) (5 2) – user98643 Oct 20 '13 at 5:27
Make all the swaps of left cycle before making swaps of the right cycle. I think the convention in (1 3 4 5) is that element in 3rd position moves to 1st position. So starting with ABCDE, (1 3 4 5) gets you CBDEA. Then, (2 3 4) gets you CDEBA. This of course is equivalent to a permutation of A,C and E and another permutation of B and D. That is, (1 3 5)(2 4). – user96614 Oct 20 '13 at 5:53

## 1 Answer

1. No, you can't switch the order, $(2\ 3\ 4)(1\ 3\ 4\ 5)\ne(1\ 3\ 4\ 5)(2\ 3\ 4)$. Cycles commute (i.e. the order of multiplication can be switched) if they are disjoint meaning no elements in common; the cycles $(1\ 3\ 4\ 5)$ and $(2\ 3\ 4)$ are not disjoint because both of them move $3$ and $4$.

2. I don't know why you wanted to switch the order. Did you think $(2\ 3\ 4)(1\ 3\ 4\ 5)$ would be easier to figure than $(1\ 3\ 4\ 5)(2\ 3\ 4)$??

3. Anyway, $(2\ 3\ 4)(1\ 3\ 4\ 5)=(1\ 4\ 5)(2\ 3)$, so you did something else wrong after switching the order. I don't know what you did; your explanation "begin with 1. 1--> 3 next 2 --> 3 and 3 --> 4. Then 2 --> 4 and 4 --> 5." makes no sense to me.

4. Here's how you multiply $(1\ 3\ 4\ 5)(2\ 3\ 4)$. Where does $1$ go? $1\to1\to3$ so $1\to3$. Where does $3$ go? $3\to4\to5$, so $3\to5$. Where does $5$ go? $5\to5\to1$, so $5\to1$. So there's one cycle, $(1\ 3\ 5)$. Now where does $2$ go? $2\to3\to4$, so $2\to4$. Finally, $4\to2\to2$, so $4\to2$. So we have another cycle, $(2\ 4)$. Putting them together, our final answer is$$(1\ 3\ 4\ 5)(2\ 3\ 4)=(1\ 3\ 5)(2\ 4)$$which, by the way is the same as $(2\ 4)(1\ 3\ 5)$ because $(1\ 3\ 5)$ and $(2\ 4)$ are disjoint cycles.

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1. I switched the order because i thought it worked like a composition, so you start from the right and move from the right cycle to the left. If (2 3 4) is A and (1 3 4 5) is B and you being with 1, B sends 1 to 3, then A sends 3 to 4. – user98643 Oct 20 '13 at 5:36
It is composition, and you do it from right to left. If (2 3 4)=A and (1 3 4 5)=B, and if you want to compute the product (i.e. composition) AB=(2 3 4)(1 3 4 5) [instead of the original problem BA=(1 3 4 5)(2 3 4)], then yes, first B sends 1 to 3, then A sends 3 to 4, so AB sends 1 to 4. But that's not what you did, is it? Your answer (1 3 4)(2 5) sends 1 to 3, not to 4. – bof Oct 20 '13 at 5:50