# How Do Permutations and Conjugacy Classes Operate in Symmetric Groups?

• Oxymoron
In summary, the congujacy class of a permutation \sigma \in S_n consists of all permutations in S_n of the same cycle type as \sigma.
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 $S_5$

(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:
(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:
(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

(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.

Does anyone know if this is right?

## What is a cycle in permutations?

A cycle in permutations is a particular type of permutation in which the elements are rearranged in a circular fashion. This means that every element is moved to the position of the next element in the cycle, and the last element is moved to the position of the first element.

## How do you find the order of a permutation?

The order of a permutation is the number of times the permutation needs to be applied to return to the original arrangement of elements. To find the order, you can write the permutation as a product of cycles and then take the least common multiple of the lengths of the cycles.

## What is a fixed point in a permutation?

A fixed point in a permutation is an element that stays in the same position after the permutation is applied. In other words, the element is unchanged by the permutation.

## What are some real-world applications of permutations?

Permutations have many practical applications, such as in computer algorithms, cryptography, and genetics. They are also used in various fields of mathematics, including combinatorics, group theory, and statistics.

## How are permutations and combinations related?

Permutations and combinations are related in that they both involve the arrangement or selection of objects. However, permutations take into account the order of the objects, while combinations do not. Permutations also include all possible arrangements, while combinations only include a subset of arrangements.

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