0
$\begingroup$

I have some questions, for the first question I will use an example:

According to my textbook:

I (1 2) (2 3) = (1 2 3)

II (2 3) (1 2) = (1 3 2)

I is not the same as II

But is I = (2 3 1) ?

The second question is whether there is an algorithm when multiplying two non disjunctive cycles. My idea is to always look at the two last cycles, the last (the right one) is always the dominant cycle. Thus I look which elements are not in the last cycle but in the first cycle and which element is on their right side, if the element in question is the last element of the cylce I look at the first element. The relation is fixed now. The element can only point at another element. I continue with the second cycle if a conflict on the assignment of the elements arrises I always prefer the assignments that I made prior. The element in question of the dominant cycle would point at the element of the first cycle which points at the element right of the element in question or if the element in question is the last element of the cylcle on the first element of the dominant cycle: I will illustrate my idea on annother example, if my idea is wrong then the example is also wrong:

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

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

If my idea is right I would like to know what is the theory behind it or if it is wrong the alternative solution and explanation.

Thank you for taking your time

$\endgroup$
0
$\begingroup$

It seems your textbook is using the right-to-left multiplication convention, so that you read the permutations from the right to the left much like you would with any other composition of maps.

As Lord Shark the Unknown has written, it is correct that $$ (1 \quad 2 \quad 3) = (2 \quad 3 \quad 1), $$ as these are both the cycles that map $$ 1 \mapsto 2, \quad 2 \mapsto 3, \quad 3 \mapsto 1. $$ The third equivalent cycle is $(3 \quad 1 \quad 2)$.

Now in the multiple $$ (1 \quad 2)(2 \quad 3), $$ we read the cycles from left to right and make a note of where each element is mapped. Let us write $\sigma = (1 \quad 2)$ and $\gamma = (2 \quad 3)$, so we are finding the multiple $\sigma\gamma$, $$ \stackrel{\sigma}{(1 \quad 2)}\stackrel{\gamma}{(2 \quad 3)}. $$ We also note that since $\gamma = (2 \quad 3)$ has no mention of $1$, it must fix $1$, so that $1 \mapsto 1$ under $\gamma$, and similarly $3 \mapsto 3$ under $\sigma$. We observe that $$ 1 \stackrel{\gamma}{\mapsto} 1 \stackrel{\sigma}{\mapsto} 2 \implies 1 \stackrel{\sigma\gamma}{\mapsto} 2. $$ Similarly $$ 2 \stackrel{\gamma}{\mapsto} 3 \stackrel{\sigma}{\mapsto} 3 \implies 2 \stackrel{\sigma\gamma}{\mapsto} 3, \quad 3 \stackrel{\gamma}{\mapsto} 2 \stackrel{\sigma}{\mapsto} 1 \implies 3 \stackrel{\sigma\gamma}{\mapsto} 1. $$ Hence $$ 1 \stackrel{\sigma\gamma}{\mapsto} 2, \quad 2 \stackrel{\sigma\gamma}{\mapsto} 3, \quad 3 \stackrel{\sigma\gamma}{\mapsto} 1 $$ is written in the cycle notation as $$ \sigma\gamma = (1 \quad 2 \quad 3). $$ I will leave it to you to confirm that $$ (2 \quad 3)(1 \quad 2) = (1 \quad 3 \quad 2). $$


As for larger compositions, we apply the same method. For example, in the product $$ (1 \quad 2)(2 \quad 3)(3 \quad 4) $$ you can do it in parts the way that you have, but a (maybe simpler) way to do it is the following. Let us write $$ \sigma = (1 \quad 2), \quad \gamma = (2 \quad 3), \quad \tau = (3 \quad 4) $$ so that we are finding the product $\sigma\gamma\tau$, $$ \stackrel{\sigma}{(1 \quad 2)}\stackrel{\gamma}{(2 \quad 3)}\stackrel{\tau}{(3 \quad 4)}. $$ Also remember that, for example, since both $3$ and $4$ are not mentioned in $\sigma = (1 \quad 2)$, we have that $3 \stackrel{\sigma}{\mapsto} 3$ and $4 \stackrel{\sigma}{\mapsto} 4$. We evaluate the product reading from right to left and check where each element goes, so that $$ 1 \stackrel{\tau}{\mapsto} 1 \stackrel{\gamma}{\mapsto} 1 \stackrel{\sigma}{\mapsto} 2 \implies 1 \stackrel{\sigma\gamma\tau}{\mapsto} 2. $$ Similarly $$ \begin{array}{c} 2 \stackrel{\tau}{\mapsto} 2 \stackrel{\gamma}{\mapsto} 3 \stackrel{\sigma}{\mapsto} 3 \implies 2 \stackrel{\sigma\gamma\tau}{\mapsto} 3 \\ 3 \stackrel{\tau}{\mapsto} 4 \stackrel{\gamma}{\mapsto} 4 \stackrel{\sigma}{\mapsto} 4 \implies 3 \stackrel{\sigma\gamma\tau}{\mapsto} 4 \\ 4 \stackrel{\tau}{\mapsto} 3 \stackrel{\gamma}{\mapsto} 2 \stackrel{\sigma}{\mapsto} 1 \implies 4 \stackrel{\sigma\gamma\tau}{\mapsto} 1. \end{array} $$ Thus $$ \sigma\gamma\tau = (1 \quad 2)(2 \quad 3)(3 \quad 4) = (1 \quad 2 \quad 3 \quad 4). $$ You can similarly show that $$ \tau\gamma\sigma = (3 \quad 4)(2 \quad 3)(1 \quad 2) = (1 \quad 4 \quad 3 \quad 2). $$

$\endgroup$
1
$\begingroup$

Yes, $(1\,2\,3)$ is the same permutation as $(2\,3\,1)$. It's the permutation that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$.

$\endgroup$
  • $\begingroup$ Can you tell me if these two equations are also correct? $\endgroup$ – RM777 Jul 5 '18 at 8:02
  • $\begingroup$ (1 2) (2 3) (3 4) = (1 2) (2 3 4) = (1 2 3 4) ) $\endgroup$ – RM777 Jul 5 '18 at 8:02
  • $\begingroup$ (3 4) (2 3) (1 2) = (3 4) (1 3 2) = (1 4 3 2) $\endgroup$ – RM777 Jul 5 '18 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.