1125: 2 Possible Meanings of Permutation of Sequence

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description/proof of 2 possible meanings of permutation of sequence

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of permutation of sequence.

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Target Context

• The reader will have a description and a proof of the 2 possible meanings of permutation of sequence.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$\mathbb{N}$:
$S$: $\subseteq \mathbb{N}$
$f$: $\in \{\text{ the sequences from }S\}$
$\sigma$: $:S\to S$, $\in \{\text{ the bijections }\}$
$\sigma (f)$: $=f\circ \sigma$
$g$: $=f\circ {\sigma }^{-1}$
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Statements:
$\sigma (f)$ is the permutation of $f$ by $\sigma$ by a definition of permutation of sequence, but some people may understand $g$ by "permutation of $f$ by $\sigma$"
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2: Proof

Whole Strategy: Step 1: see what $\sigma (f)$ means; Step 2: see what $g$ means.

Step 1:

Let us see what $\sigma (f):=f\circ \sigma$ means.

For each $s\in S$, $\sigma (f)(s)=f(\sigma (s))$, which means that $\sigma (f{)}_j=f(\sigma (s_j))=f({\sigma }_j)$ where $s_j\in S$ is the $j$-th element of $S$, which means that the sequence is $f({\sigma }_1),f({\sigma }_2),...$.

Note that when we regard the permutation as moving an item to a position, the permutation is moving the ${\sigma }_j$ item to the $j$-th position.

For example, when $S=\{1,2,3\}$, $f=4,5,6$, and $\sigma :(1,2,3)\mapsto (3,1,2)$, $\sigma (f)=f(\sigma (1)),f(\sigma (2)),f(\sigma (3))=6,4,5$, which is moving the $\sigma (1)=3$ item to the $1$-st position, the $\sigma (2)=1$ item to the $2$-nd position, and the $\sigma (3)=2$ item to the $3$-rd position.

Step 2:

The reason why we have described this proposition is that Step 1 may not be what some people understand as "permutation of $f$ by $\sigma$".

They may understand it as moving the $j$-th item to the ${\sigma }_j$ position.

Let us see that that is $g$ not $\sigma (f)$.

Let us denote the permutation that moves the $j$-th item to the ${\sigma }_j$ position as $h$.

$f(s_j)=h({\sigma }_j)=h(\sigma (s_j))$.

Taking $t:=\sigma (s_j)$, $s_j={\sigma }^{-1}(t)$, so, $h(t)=f(s_j)=f({\sigma }^{-1}(t))=f\circ {\sigma }^{-1}(t)$.

So, $h=f\circ {\sigma }^{-1}=g$ not $=f\circ \sigma :=\sigma (f)$.

For example, when $S=\{1,2,3\}$, $f=4,5,6$, and $\sigma :(1,2,3)\mapsto (3,1,2)$, which is the same with the above example, ${\sigma }^{-1}:(1,2,3)\mapsto (2,3,1)$ and $g=f({\sigma }^{-1}(1)),f({\sigma }^{-1}(2)),f({\sigma }^{-1}(3))=5,6,4$, which is moving the $1$-st item to the $3$-rd position, the $2$-nd item to the $1$-st position, and the $3$-rd item to the $2$-nd position.

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3: Note

This is not about which of the 2 is correct, but about the necessity of clarifying what is meant by 'permutation of $f$ by $\sigma$' and being consistent.

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References

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