How Do Permutations and Conjugacy Classes Operate in Symmetric Groups?

[Oxymoron]

Question

Let S_n be the symmetric group on n letters.

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

\phi\sigma\phi^{-1} = (\phi(x_1),\dots,\phi(x_k))

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

(iii) In the case of S_5, give the numbers of permutations of each cycle type

(iv) Find all normal subgroups of S_5

 

[Oxymoron]

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

\phi\sigma\phi^{-1} = (\phi(x_1),\dots,\phi(x_k))

 

[Oxymoron]

(iii) The list of cycle types in S_5 and the number of permutations for each:
(1) \equiv [1^5] = 5!/(1^5*5!) = 1
(1 2) \equiv [1^3*2*3!) = 5!/(1^3*2*3!) = 10
(1 2)(3 4) \equiv [1^1*2^2] = 5!/(1^1*2^2*2!) = 15
(1 2 3) \equiv [1^1*3^1] = 5!/1^2*3*2!) = 20
(1 2 3)(4 5) \equiv [2*3] = 5!/(2*3) = 20
(1 2 3 4) \equiv [1^1*4^1] = 5!/(1*4) = 30
(1 2 3 4 5) \equiv [5] = 5!/5 = 24

And 1 + 10 + 15 + 20 + 20 + 30 + 24 = 120 = 5!

 

[Oxymoron]

(iv) A subgroup H of the symmetric group S_n is normal if \phi H = H\phi for some \phi \in S_n. Equivalently, if \phi H \phi^{-1} = H for all \phi \in S_n. That is, H is a normal subgroup of S_n if and only if, each conjugacy class of S_n is either entirely inside H or outside H.

But we know \phi H\phi^{-1} = \phi H hence \phi is the kernel of the homomorphism

 

[Oxymoron]

(ii) Let \alpha \in S_n be of the same cycle type as \sigma \in S_n. Define \phi \in S_n to be that permutation which maps each element x_i in the cycle of \sigma to the corresponding a in the corresponding cycle of \alpha. In other words, \phi : \sigma(x_i) \rightarrow \alpha(a_i). Which is equivalent in saying that \phi(\sigma(x_i)) = \alpha(a_i). Therefore, from the first part of the question we have

\alpha = \phi\sigma\phi^{-1}

Therefore every pair of elements with the same type are conjugate.

 

[Oxymoron]

Does anyone know if this is right?