0
$\begingroup$

Checking my work.

In either direction:

$(1243)[1] = 2$ and $(5)[2] = 2$, so far we have $(1, 2,\ldots$

$(1243)[2] = 4$ and $(5)[4] = 4$, so far we have $(1, 2, 4,\ldots$

$(1243)[4] = 3$ and $(5)[3] = 3$, so far we have $(1, 2, 4, 3, \ldots$

$(1243)[3] = 1$ and $(5)[1] = 1$, so we have $(1, 2, 4, 3).$

$(1243)[5] = 5$ and $(5)[5] = 5$, so we have $(5)$.

So, $(1243)(5) = (24135)$.

$(5)[1] = 1$ and $(1243)[1] = 2$, so far we have $(1, 2,\ldots$

$(5)[2] = 2$ and $(1243)[2] = 4$, so far we have $(1, 2, 4,\ldots$

$(5)[4] = 4$ and $(1243)[4] = 3$, so far we have $(1, 2, 4, 3,\ldots$

$(5)[3] = 3$ and $(1243)[3] = 1$, so we have $(1, 2, 4, 3)$.

$(5)[5] = 5$ and $(1243)[5] = 5$, so we have $(5)$.

So, $(1243)(5) = (24135)$.

Does that make sense?

edit:

I am trying to organize the topic of permutaions in my head. Here's what I wrote down(the question stems from it). Please, see what I need to correct:

Ways to represent permutations:

a) In ordered pairs form: $f = \{(x_1 y_1), (x_2 y_2), \ldots, (x_n y_n)\}$.

b) In $2 \times n $ matrix form: $f = \begin{pmatrix} x_1 & x_2 & \ldots & x_n \\ y_1 & y_2 & \ldots & y_n \end{pmatrix}$.

c) In cyclic form: $f = (x_1, f(x_1), f^2(x_1), \ldots, f^n(x_1)).$

d) In the form of values of the range of $f: f = (y_1, y_2, \ldots, y_n)$.

Example: $f = \underbrace{\{(12)(24)(31)(43)(55)\}}_{\text{(a)}} = \underbrace{\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 1 & 3 & 5 \end{pmatrix}}_{\text{(b)}} = \underbrace{(1243)(5)}_{\text{(c)}} = \underbrace{(24135)}_{\text{(d)}}$

$\endgroup$
11
  • 1
    $\begingroup$ Are these meant to be cyclic permutations? - I'm a little confused about what we're working with here. $\endgroup$
    – Alekos Robotis
    Jul 25, 2015 at 22:41
  • $\begingroup$ Yes, these are cyclic permutations. $\endgroup$
    – number
    Jul 25, 2015 at 22:42
  • 1
    $\begingroup$ What do you mean by $(5)$? The convention with disjoint cycle notation is that we ignore $1$-cycles - so $(5)=(1)(2)(3)(4)(5)=\text{id}$. $\endgroup$
    – John Gowers
    Jul 25, 2015 at 22:43
  • $\begingroup$ no. since (1243)(5)[5]=5 and (24135)[5]=2 $\endgroup$
    – sha
    Jul 25, 2015 at 22:43
  • $\begingroup$ If this asks what I think it does, the answer to your question is $(1243) \circ (5) = (1243)$. This is because permutations of a single element to itself are trivial permutations. $\endgroup$
    – Alekos Robotis
    Jul 25, 2015 at 22:44

1 Answer 1

Reset to default
2
$\begingroup$

As written, the question asks for the result of $(1243) \circ (5)$. However, this is appears to be trivial, because $(5)$ just means an identity mapping, mapping $5 \rightarrow 5$, the empty permutation. In other words, if we first apply $(5)$, and then $(1243)$, $(5)$ does nothing, leaving $(1243) \circ (5) = (1243)$.

As far as I can see, your mistake seems to be a misunderstanding of cyclic permutation notation: $(1243)$ is the permutation $1\rightarrow2\rightarrow4\rightarrow3\rightarrow1$. $(5)$ implies $5 \rightarrow 5$.

So, if we tried to set $(1243)(5)=(24135)$, we would get $1\rightarrow2\rightarrow4\rightarrow3\rightarrow1$ is equivalent to $2\rightarrow4\rightarrow1\rightarrow3\rightarrow5\rightarrow2$. This is not true.

$\endgroup$
7
  • $\begingroup$ In the OP, do you think $(c) = (d)$ (not the example)? $\endgroup$
    – number
    Jul 25, 2015 at 22:58
  • $\begingroup$ There is a problem: your $2$ by $n$ matrix is a matrix depicting a series of disjoint cycles. That is $x_1 \rightarrow y_1, ... , x_n\rightarrow y_n$. What I mean to say is that your example differs fundamentally from your $4$ ways to represent cyclic permutations, and those do not all represent the same things, as written. $\endgroup$
    – Alekos Robotis
    Jul 25, 2015 at 23:03
  • $\begingroup$ For your example, the first three parts make sense, but the third and fourth are not equivalent. $\endgroup$
    – Alekos Robotis
    Jul 25, 2015 at 23:06
  • $\begingroup$ How about $f = \begin{pmatrix} x_1 & y_1 & \ldots & y_n \\ y_1 & y_2 & \ldots & y_{n + 1} \end{pmatrix} = (x_1, y_1, y_2, \ldots, y_{n + 1}) = (y_1, y_2, \ldots, y_{n + 1})$. $\endgroup$
    – number
    Jul 25, 2015 at 23:13
  • $\begingroup$ Your first step is correct, however the second step is not. These cycles can not be equal, they don't contain the same elements. That is, the right hand side $(y_1, y_2,...,y_{n+1}) = (x_1)(y_1, y_2,...,y_{n+1})$. This is not equal to the left hand side. $\endgroup$
    – Alekos Robotis
    Jul 25, 2015 at 23:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .