**trajectories** of the permutation of $S_6$ , $\sigma = (1 2 356) (345)$

Question

I have been asked to find the trajectories of the permutation of $S_6$ , $\sigma = (1 2 356) (345)$ and I have to write $\sigma$ as the product of discrete cycles.

Can anyone please tell me what are the meanings of the bold words ?

Answer 1

In this question they call it the orbit, which is I think what's meant: what is the image of $i \in \{1,2,3,4,5,6\}$ under successive applications of $\sigma$? So the orbit of $1$ is $\{1,2,3,4,6\}$ (this is also the orbit of all $i \neq 5$ of course) and the orbit of $5$ is $\{5\}$.

For the second: it's a standard fact that any permutation can be canonically written as a product of cycles such that different cycles use different numbers; normally called disjoint cycles.

If you look at $\sigma$, $1 \to 2 \to 3 \to 4 \to 6 \to 1$ and $5$ is a fixpoint. So we can write $\sigma$ simply as $(1 2 3 4 6)(5)$ (we can omit the $(5)$) and its powers are then easy to write down..

Both tasks are closely related as you can see.

Answer 2

The decomposition of a permutation $\sigma\in S_n$ into disjoint ("discrete"(?)) cycles pops up when you consider the natural action of $\langle\sigma\rangle$ on $X=\{1,\dots,n\}$ (i.e. the action as a group of permutations). In fact, this action partitions $X$ into orbits ("trajectories"(?)), say $X=\bigsqcup_{i=1}^NO_i$, and the disjoint cycles are the permutations $\alpha_i\in S_n$ defined, for each $i=1,\dots,N$, as the extension to $X\setminus O_i$ by the map $Id_{X\setminus O_i}$, of the restriction $\sigma_{|O_i}$:

\begin{alignat}{1} &\alpha_i(k)=\sigma(k), \space\text{for} \space k\in O_i \\ &\alpha_i(k)=k, \space\text{for} \space k\in O_{j\ne i} \\ \end{alignat}

The "disjoint" refers to the fact that $\operatorname{supp}(\alpha_i)\cap \operatorname{supp}(\alpha_j)=\emptyset$, for every $i,j\in \{1,\dots,N\}, i\ne j$, where $\operatorname{supp}(\alpha_i):=\{k\in X\mid \alpha_i(k)\ne k\}$.

The following deploys this general framework into your particular case of $\sigma=(12356)(345)\in S_6$.

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The permutation you are given is, in matrix notation, $\sigma=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 6 & 5 & 1 \end{pmatrix}$. Then:

$$\sigma^2=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 4 & 6 & 1 & 5 & 2 \end{pmatrix}$$

$$\sigma^3=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 6 & 1 & 2 & 5 & 3 \end{pmatrix}$$

$$\sigma^4=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 1 & 2 & 3 & 5 & 4 \end{pmatrix}$$

$$\sigma^5=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \end{pmatrix}=()$$

Now, consider the natural action of $\langle\sigma\rangle$ on $X=\{1,2,3,4,5,6\}$. What you call the "trajectories of $\sigma$" are most probably the orbits under such a $\langle\sigma\rangle$-action on $X$, namely:

\begin{alignat}{1} O(1) &=\{\sigma^k(1), k=1,2,3,4,5\} = \{2,3,4,6,1\} \\ O(2) &=\{\sigma^k(2), k=1,2,3,4,5\} = \{3,4,6,1,2\}=O(1) \\ O(3) &=\{\sigma^k(3), k=1,2,3,4,5\} = \{4,6,1,2,3\}=O(2)=O(1) \\ O(4) &=\{\sigma^k(4), k=1,2,3,4,5\} = \{6,1,2,3,4\}=O(3)=O(2)=O(1) \\ O(5) &=\{\sigma^k(5), k=1,2,3,4,5\} = \{5\} \\ O(6) &=\{\sigma^k(6), k=1,2,3,4,5\} = \{1,2,3,4,6\}=O(4)=O(3)=O(2)=O(1) \\ \end{alignat}

So, there are:

• one "trajectory" of size $5$, $O_1=\{1,2,3,4,6\}\space(=O(6)=O(4)=O(3)=O(2)=O(1))$;
• one "trajectory" of size $1$, $O_2=\{5\}\space(=O(5))$.

Now, consider the permutations $\alpha_1,\alpha_2\in S_6$ defined by: $\alpha_i(k)=\sigma(k)$ for $k\in O_i$, and $\alpha_i(k)=k$ for $k\notin O_i$, $i=1,2$. You can verify that $\sigma=\alpha_1\alpha_2$ and $\operatorname{supp}(\alpha_1)\cap \operatorname{supp}(\alpha_2)=\emptyset$, where, for any $\tau\in S_6$, $\operatorname{supp}(\tau):=\{k\in X\mid \tau(k)\ne k\}$. Such $\alpha_1,\alpha_2$ are the disjoint ("discrete"(?)) cycles which $\sigma$ can be split into. According to this definition, for your $\sigma$ we get $\sigma=(12346)(5)=(12346)$.