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I generally do not have any problem multiplying cycles, but I've seen on Wikipedia that

$$(2 5) = (2 3) (3 4) (4 5) (4 3) (3 2). $$

I started following the path of $2$ on the right:

$$2\to3\to4\to5\to \ ?$$

Where does $5$ go? I should stop here, right? Then $2\to 5$, that is, $(25)$. But what about $(23)(34)$?

Daniel W. Farlow
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Guerlando OCs
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    When a cycle does not contain an element, that element stays there. –  Aug 03 '15 at 04:36
  • $5 \rightarrow 5 \rightarrow 5.$ This says 2 goes to 5. you aslo need to check where 5 goes! – Makarand Sarnobat Aug 03 '15 at 04:38
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    $$\begin{array}{l}\phantom{=} ,,(23)(34)(45)(43)(32)2 \ =(23)(34)(45)(43)3 \ = (23)(34)(45)4 \ = (23)(34)5 \ =(23)5 \ = 5. \end{array} $$ Notice how $(34)5=5$ and $(23)5=5$. By definition, $(34)$ swaps $3$ and $4$ and does nothing to the other numbers. So, $(34)5=5$, and similarly $(23)5=5$. – anon Aug 03 '15 at 04:53
  • The key point @frank000 is referring to is that, more explicitly, when a cycle does not contain an element, such an element maps to itself. That is, suppose a cycle consist of $a,b,c,d$, and you have a product like $(a;b;c)(d)$. How do you compute such a product? Starting off, $d\mapsto ?$; well, clearly, $d$ maps to itself for $(d)$. Where does $d$ map to in $(a;b;c)$? It maps to itself, and that was, I believe, the point of coldnumber using the two-row form: to make it even more explicit that $5$ is mapping to itself. This fact is more apparent with two-row notation than cycle. – Daniel W. Farlow Aug 03 '15 at 05:06

4 Answers4

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This is another conjugation problem in disguise:

$(2\ 3)(3\ 4)(4\ 5)(4\ 3)(3\ 2) = (2\ 3)[(3\ 4)(4\ 5)(3\ 4)^{-1}](3\ 2)$

$= (2\ 3)(3\ 5)(2\ 3)^{-1}$ (since $(3\ 4)$ takes $4 \to 3$ and fixes $5$)

$= (2\ 5)$ (since $(2\ 3)$ takes $3 \to 2$ and fixes $5$).

David Wheeler
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Let's go through the product(s) $$ (2 3) (3 4) (4 5) (4 3) (3 2) $$ step by step.

Start with $2$, by writing $(2$, and see how the elements map out:

  • $2\mapsto 3\mapsto 4\mapsto 5\mapsto 5\mapsto 5\qquad:\qquad(2\; 5$
  • $5\mapsto 5\mapsto 5\mapsto 4\mapsto 3\mapsto 2\qquad:\qquad(2\; 5)(3$
  • $3\mapsto 2\mapsto 2\mapsto 2\mapsto 2\mapsto 3\qquad:\qquad(2\; 5)(3)(4$
  • $4\mapsto 4\mapsto 3\mapsto 3\mapsto 4\mapsto 4\qquad:\qquad(2\;5)(3)(4)$

The process has terminated and we know $(2\;5)(3)(4)=(2\;5)$. That is, we can see that $$ (2\;5) = (2 3) (3 4) (4 5) (4 3) (3 2). $$ Did all of those steps make sense?

Daniel W. Farlow
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Assuming there are cycles in $S_5$, $$(2\ 3) = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 2 & 4 & 5 \end{pmatrix}=(2\ 3)(1)(4)(5) \\ (3\ 4) = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 4 & 3 & 5 \end{pmatrix}=(3\ 4)(1)(2)(5)$$

Hence, $5\mapsto 5$ in both of those cycles.

However, to conclude that $5 \mapsto 2$ you need to follow it through the cycles as you followed 2, starting from the right to the left.

See my answer here for a detailed explanation of composing cycles: Shortcut for composing cycles

Community
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coldnumber
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Starting from the right, we see that $2\to3\to4\to5$ and $5\to4\to3\to2$. On the other hand, $3\to2\to3$. Every other number just maps to itself, which you can check for yourself.

Note that if a number is fixed, then we don't write it in the shorthand notation for the permutation. Thus we just have $(2\ 5)$

pancini
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