description/proof of that for n-symmetric group and n-cycle, centralizer of cycle on symmetric group is cyclic group by cycle
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of n-symmetric group.
- The reader knows a definition of m-cycle on n-symmetric group.
- The reader knows a definition of centralizer of element on group.
- The reader knows a definition of cyclic group by element.
Target Context
- The reader will have a description and a proof of the proposition that for any n-symmetric group and any n-cycle, the centralizer of the cycle on the symmetric group is the cyclic group by the cycle.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(S_n\): \(= \text{ the } n \text{ -symmetric group }\)
\(p\): \(= (\sigma_1, ..., \sigma_n)\), \(\in \{\text{ the } n \text{ -cycles on } S_n\}\)
\(C_G (p)\): \(= \text{ the centralizer of } p \text{ on } G\)
\(\langle p \rangle\): \(= \text{ the cyclic group by } p\)
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Statements:
\(C_G (p) = \langle p \rangle\)
//
2: Natural Language Description
For the \(n\)-symmetric group, \(S_n\), any \(n\)-cycle on \(S_n\), \(p = (\sigma_1, ..., \sigma_n)\), the centralizer of \(p\) on \(G\), \(C_G (p)\), and the cyclic group by \(p\), \(\langle p \rangle\), \(C_G (p) = \langle p \rangle\).
3: Proof
Whole Strategy: Step 1: see that for each element of \(C_G (p)\), the element is the identity element or does not preserve any element of \(\{1, ..., n\}\); see that the element that does not preserve any element of \(\{1, ..., n\}\) shifts \(\{\sigma_1, ..., \sigma_n\}\) by a constant; Step 3: see that \(\langle p \rangle \subseteq C_G (p)\).
Let \([j]\) denote the residue of \(j\) divided by \(n\) with \(0\) replaced by \(n\).
So, \(1 \le [j] \le n\).
For example, when \(n = 2\), \([1] = 1, [2] = 2, [3] = 1, ...\).
The purpose is to avoid having to say like "when \(j + k \le n\), \(j + k\), but otherwise, \(j + k - n m\) such that \(1 \le j + k - n m \le n\)"; instead, we can just say \([j + k]\).
Someone may want to say that that is just a modulo \(n\), and in idea, that is so, but strictly speaking, the modulo of \(1\) by \(2\) is not \(1\) but the equivalence class of \(1\), and so, we do not say so.
Step 1:
Let us see that for each element of \(C_G (p)\), the element is the identity element or does not preserve any element of \(\{1, ..., n\}\).
Let us suppose that a \(p' \in C_G (p)\) preserves \(\sigma_j\).
\(p' p p'^{-1} = p\), so, \(p' p = p p'\).
\(p' p (\sigma_j) = p' (\sigma_{[j + 1]}) = p p' (\sigma_j) = p (\sigma_j) = \sigma_{[j + 1]}\). That means that \(p'\) preserves also \(\sigma_{[j + 1]}\). Then, \(p'\) preserves also \(\sigma_{[j + 2]}, ..., \sigma_{[j + n - 1]}\), which means that \(p'\) preserves all the elements of \(\{\sigma_1, ..., \sigma_n\}\), so, \(p' = id\).
Step 2:
Let us \(p' \in C_G (p)\) be any that does not preserve any element of \(\{1, ..., n\}\).
\(p' p (\sigma_j) = p' (\sigma_{[j + 1]}) = \sigma_{[j + 1 + k]}\) for a \(k \in \mathbb{N} \setminus \{0\}\) such that \(1 \le k \lt n\).
\(p p' (\sigma_j) = p (\sigma_{[j + l]}) = \sigma_{[j + l + 1]}\) for an \(l \in \mathbb{N} \setminus \{0\}\) such that \(1 \le l \lt n\).
\(\sigma_{[j + 1 + k]} = \sigma_{[j + l + 1]}\), which implies that \(k = l\). So, \(p'\) shifts \(\sigma_{[j + 1]}\) and \(\sigma_j\) by \(k\).
Likewise, \(p'\) shifts \(\sigma_{[j + 2]}\) and \(\sigma_{[j + 1]}\) by \(k'\), but \(k' = k\), because \(\sigma_{[j + 1]}\) is already known to be shifted by \(k\).
So, \(p'\) shifts \(\{\sigma_j, ..., \sigma_{[j + n - 1]}\} = \{\sigma_1, ..., \sigma_n\}\) by \(k\).
That means that \(p' = (\sigma_1, ..., \sigma_n)^k\).
So, \(C_G (p) \subseteq \langle p \rangle\).
Step 3:
Let us see that \(\langle p \rangle \subseteq C_G (p)\).
\(p^k p (p^k)^{-1} = p^k p p^{-k} = p^{k + 1} p^{-k} = p\).
So, \(C_G (p) = \langle p \rangle\).
