# Homework Help: Cycles and Permutations

1. Apr 23, 2005

### 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 $S_5$

2. Apr 23, 2005

### 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

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

Last edited: Apr 24, 2005
3. Apr 24, 2005

### 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!

Last edited: Apr 24, 2005
4. Apr 24, 2005

### 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

5. Apr 24, 2005

### 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.

6. Apr 25, 2005

### Oxymoron

Does anyone know if this is right?