1
$\begingroup$

If $\alpha=(1234)$, $\{1,3\}\alpha=\{1\alpha,3\alpha\}=\{2,4\}$ and $\{2,4\}\alpha=\{1,3\}$. If $\beta=(123)$, $\{1,3\}\beta=\{2,1\}$ and $\{2,4\}\beta=\{4,3\}$.

$\alpha$ partitions the set $\{1,2,3,4\}$ into the subsets $\{1,3\}$ and $\{2,4\}$ but $\beta$ does not.


As $\beta$ is a permutation, it is a bijective function. $\beta=(123)$ and is not defined for 4. Why does $4\beta=4$?

Is $4\beta$ here treated as $(4)(123)$?

  • 3
    Yes it is use to omit the one-cycles!2017-02-15
  • 0
    @PeterMelech Thank you for the quick clarification.2017-02-15
  • 0
    You´re welcome!2017-02-15

1 Answers 1

2

Each permutation $\pi\in\mathscr{S}_n$ is a bijection to and from the set $\overline{1,n}:=\{1,\dots n\}$.

When $n=4$, on one hand, we can clearly see that $\alpha=(1234)$ is a bijection to and from $\overline{1,4}$; explicitly:

$$\begin{align} \alpha: \{1,2,3,4\} &\to \{1,2,3,4\} \\ 1 &\mapsto 2, \\ 2 &\mapsto 3, \\ 3 &\mapsto 4, \\ 4 &\mapsto 1. \end{align}$$

On the other hand, still with $n=4$, we have $\beta=(123)$; in this context, $\beta\stackrel{!}{=}(123)\color{red}{(4)}$ is defined by

$$\begin{align} \beta: \{1,2,3,4\} &\to \{1,2,3,4\} \\ 1 &\mapsto 2, \\ 2 &\mapsto 3, \\ 3 &\mapsto 1, \\ \color{red}{4} &\,\color{red}{\mapsto 4}, \end{align}$$ because the $\color{red}{\text{one-cycle } (4)}$ is inferred from the context.

One-cycles are typically omitted for convenience . . .

But they're still there.

If we were to consider a bigger value of $n$, like $n=5$, say, then the $\alpha$ and $\beta$, taken as elements of $\mathscr{S}_4$, strictly speaking, are not elements of $\mathscr{S}_5$ (although, of course, they each define a bijection from $\overline{1,5}$ to $\overline{1,5}$; for instance: define $\alpha':\overline{1,5}\to\overline{1,5}$ by $\color{blue}{\alpha'(5)=5}$ and $\alpha'(x)=\alpha(x)$ for all $x\in\overline{1,4}$, so that $\alpha'=(1234)\color{blue}{(5)}\neq\alpha$, even though, if the context is clear enough, we may write $\alpha'=(1234)$ sometimes).