Let be $\sigma=(1\cdots n) \in S_n$. Then $C_{S_{n}}(\sigma) = \langle \sigma \rangle$.

Question

Beggining of the proof: Note that every power of $\sigma$ comutes with $\sigma$, then $\langle \sigma \rangle \subset C_{S_{n}}(\sigma)$.

I have two questions:

1. Why $|\langle \sigma \rangle|= o(\sigma)=n$ ?

2. Why is enough show that $|C_{S_{n}}(\sigma)|=n$ (after show item 1) to prove that $C_{S_{n}}(\sigma) \subset \langle \sigma \rangle$?

Answer 1

1. By definition, $\langle \sigma \rangle$ is the smallest subgroup of $S_n$ that contains $\sigma$. We can show that $\langle \sigma \rangle = \{\sigma^n \vert 0 \leq n < d\}$. Firstly, it contains the identity element since $e = \sigma^0$. For inverse elements and closure under multiplication consider the following:

Let $d := \vert \sigma \vert$ the order of the element $\sigma$. Then we can write any $n \in \mathbb{N}$ as $n = m+cd$ where $m,c \in \mathbb{N}$. Then we have

$$\sigma^n = \sigma^{m + cd} = \sigma^m$$

So in particular for any $\sigma^n, \sigma^m \in \{\sigma^n \vert 0 \leq n < d\}$ we have that $\sigma^n \sigma^m = \sigma^{n+m} = \sigma^{k + cd} = \sigma^k$ with $k < d$.

For the inverse, note that for any $\sigma^m \in \{\sigma^n \vert 0 \leq n < d\}$ we have $\sigma^m \sigma^{d-m} = \sigma^d = \sigma^0 = e$. Hence $\{\sigma^n \vert 0 \leq n < d\}$ indeed forms a subgroup and it contains $\sigma$. By the construction we can also see that it is the smallest with this property.

2. Now if we can show that $C_{S_n}(\sigma)$ has cardinality $d$, and $\langle \sigma \rangle$ has cardinality $d$, and we know that $\langle \sigma \rangle \subseteq C_{S_n}(\sigma)$ then we know that $d$ of the $d$ elements in $C_{S_n}(\sigma)$ are the ones in $\langle \sigma \rangle$ i.e. the two are equal.

Answer 2

1)

For every group $G$ and every element $g\in G$ we have: $$<g>=\{g^k\mid k\in\mathbb Z\}$$If $g$ has finite order $n$ then this subgroup takes the shape:$$\{1,g,\dots,g^{n-1}\}$$so that its cardinality is $n$.

2)

$|C_{S_n}(\sigma)|=n$ tells us that next to $1,\sigma,\dots,\sigma^{n-1}$ there are no other elements of $S_n$ that belong to $C_{S_n}(\sigma)$.

Answer 3

An $n$- cycle always has order $n$.

For the second part, one can use the orbit stabilizer theorem to get that $|C_{S_n}(\sigma)|=n$. (Obviously $\langle\sigma\rangle\subset C_{S_n}(\sigma)$ ). But, when a subset of a finite set has the same order as the set, the two are equal.