Cycles and Permutations

  • Thread starter Oxymoron
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  • #1
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Question

Let [itex]S_n[/itex] be the symmetric group on [itex]n[/itex] letters.

(i) Show that if [itex]\sigma = (x_1,\dots,x_k)[/itex] is a cycle and [itex]\phi \in S_n[/itex] then

[tex]\phi\sigma\phi^{-1} = (\phi(x_1),\dots,\phi(x_k))[/tex]

(ii) Show that the congujacy class of a permutation [itex]\sigma \in S_n[/itex] consists of all permutations in [itex]S_n[/itex] of the same cycle type as [itex]\sigma[/itex]

(iii) In the case of [itex]S_5[/itex], give the numbers of permutations of each cycle type

(iv) Find all normal subgroups of [itex]S_5[/itex]
 

Answers and Replies

  • #2
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(i) Let [itex]\sigma = (x_1, \dots, x_k)[/itex]. Then since [itex]\sigma(x_i) = x_{i+1}[/itex] for [itex](1\leq i < k)[/itex] and [itex]\sigma(x_k) = x_1[/itex], then [itex]\phi\sigma\phi^{-1}(\phi(x_i)) = \phi\sigma(x_i) = \phi(x_{i+1})[/itex] and [itex]\phi\sigma\phi^{-1}(\phi(x_k)) = \phi\sigma(x_k) = \phi(x_1)[/itex]. So we can write

[tex]\phi\sigma\phi^{-1} = (\phi(x_1),\dots,\phi(x_k))[/tex]
 
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  • #3
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(iii) The list of cycle types in [itex]S_5[/itex] and the number of permutations for each:
[itex](1)[/itex] \equiv [1^5] = 5!/(1^5*5!) = 1
[itex](1 2)[/itex] \equiv [1^3*2*3!) = 5!/(1^3*2*3!) = 10
[itex](1 2)(3 4)[/itex] \equiv [1^1*2^2] = 5!/(1^1*2^2*2!) = 15
[itex](1 2 3)[/itex] \equiv [1^1*3^1] = 5!/1^2*3*2!) = 20
[itex](1 2 3)(4 5)[/itex] \equiv [2*3] = 5!/(2*3) = 20
[itex](1 2 3 4)[/itex] \equiv [1^1*4^1] = 5!/(1*4) = 30
[itex](1 2 3 4 5)[/itex] \equiv [5] = 5!/5 = 24

And 1 + 10 + 15 + 20 + 20 + 30 + 24 = 120 = 5!
 
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  • #4
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(iv) A subgroup [itex]H[/itex] of the symmetric group [itex]S_n[/itex] is normal if [itex]\phi H = H\phi[/itex] for some [itex]\phi \in S_n [/itex]. Equivalently, if [itex]\phi H \phi^{-1} = H[/itex] for all [itex]\phi \in S_n[/itex]. That is, [itex]H[/itex] is a normal subgroup of [itex]S_n[/itex] if and only if, each conjugacy class of [itex]S_n[/itex] is either entirely inside [itex]H[/itex] or outside [itex]H[/itex].

But we know [itex]\phi H\phi^{-1} = \phi H[/itex] hence [itex]\phi[/itex] is the kernel of the homomorphism
 
  • #5
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(ii) Let [itex]\alpha \in S_n[/itex] be of the same cycle type as [itex]\sigma \in S_n[/itex]. Define [itex]\phi \in S_n[/itex] to be that permutation which maps each element [itex]x_i[/itex] in the cycle of [itex]\sigma[/itex] to the corresponding [itex]a[/itex] in the corresponding cycle of [itex]\alpha[/itex]. In other words, [itex]\phi : \sigma(x_i) \rightarrow \alpha(a_i)[/itex]. Which is equivalent in saying that [itex]\phi(\sigma(x_i)) = \alpha(a_i)[/itex]. Therefore, from the first part of the question we have

[tex]\alpha = \phi\sigma\phi^{-1}[/tex]

Therefore every pair of elements with the same type are conjugate.
 
  • #6
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Does anyone know if this is right?
 

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