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Lemma on Permutations clarification

  1. Jun 22, 2014 #1
    1. The problem statement, all variables and given/known data
    If k is moved by σ, then σk is also moved by σ

    proof: otherwise σk is fixed by σ, that is σ(σk) = σk. But the fact that σ is one-to-one gives σk = k, which is contrary to the hypothesis.

    I am confused trying to understand this. I don't understand the part that says "But the fact that σ is one-to-one gives σk = k, which is contrary to the hypothesis." I am not ignorant of what one-to-one means, but am having trouble putting together why that gives σk = k.

    It seems logical to me you could have σ to a mapping that defines (1,2,3) → (4,5,6)
    where k = {1,2,3} so σk(1) = (4) and σ(σk(1) = σ(4) = not in k

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 22, 2014 #2


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    Please define your variables. My natural assumption is that [itex]k \in \{1,\dots, n\}[/itex] and [itex]\sigma : \{1,\dots,n\} \to \{1,\dots,n\}[/itex] is a permutation.

    Let [itex]\sigma(k) = j[/itex]. We have that [itex]\sigma[/itex] is one-to-one, so by definition for any [itex]i \in \{1,\dots,n\}[/itex], if [itex]\sigma(i) = j[/itex] then [itex]i = k[/itex].

    If [itex]\sigma(j) = j[/itex] then, by the above, [itex]j = k[/itex].
  4. Jun 22, 2014 #3


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    That mapping is not a permutation. You also need to define what 4, 5, and 6 map to.

    Another way to see why the proof works is to multiply by the inverse permutation. That gives σ-1σ(σk) = σ-1σk
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