The order of the element $\sigma$ of the symmetric group $S_5$ is

Question

The order of the element $\sigma = \begin{pmatrix} 1 &2 &3 &4 &5 \\ \downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\ 4 &5 &1 &3 &2 \end{pmatrix}$ of the symmetric group $S_5$ is

$\text{(A)}\space 2$ $\space\space\space\space\space\space\space\space\space\space$ $\text{(B)}\space 3$ $\space\space\space\space\space\space\space\space\space\space$ $\text{(C)}\space 6$ $\space\space\space\space\space\space\space\space\space\space$ $\text{(D)}\space 8$ $\space\space\space\space\space\space\space\space\space\space$ $\text{(E)}\space 12$

---

My ATTEMPT: Regardless of the notation I use:

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

So we required $6$ steps to reach to the identity $\begin{pmatrix} 1 &2 &3 &4 &5 \end{pmatrix}$.

Hence $\text{(C)}$ is the correct option.

---

In fact, the given answer is also $\text{(C)}$. However:

$(1)$ Is it allowed to use that notation? I never saw it before, but I saw something like

$\begin{pmatrix} a &b &c\\ \end{pmatrix}\begin{pmatrix} d &e\\ \end{pmatrix}\begin{pmatrix} f &g\\ \end{pmatrix}$, but I do not know how to convert it to be as such.

$(2)$ Getting $\text{(C)}$ does not mean my approach is right, I am not sure. Is it right?

$(3)$ Is there a quicker way to find the order?

Answer 1

$(1)$ The notation $(abc)(de)(fg)$ groups together the elements of a cycle. Hence in this notation, $\sigma = \begin{pmatrix} 1 &2 &3 &4 &5 \\ \downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\ 4 &5 &1 &3 &2 \end{pmatrix}$ would be $(1\space\space\space4\space\space\space3)(2\space\space\space5)$ (which actually means$(1\rightarrow4\rightarrow3\rightarrow1)(2\rightarrow 5\rightarrow 2)$).

$(2)$ The approach is correct.

$(3)$ Another approach might be to use the cycle notation. Since $\sigma=(1\space\space\space4\space\space\space3)(2\space\space\space5)$, and the two cycles are of order $3$ and $2$ respectively, the order of $\sigma$ would be $2\times 3=6$.

Answer 2

That is a notation which is indeed not used so first I will explain that notation to you.

Let $1\leq a_1,\dots,a_k\le n$. Define $(a_1\ a_2\ \cdots\ a_k)\in\mathrm S_n$ to be the permutation that sends $a_1$ to $a_2$, $a_2$ to $a_3$, and so on sending $a_{k-1}$ to $a_k$. Finally, it sends $a_k$ to $a_1$. Everything which is not included in the $a_i$s goes to itself. As an example, the permutation $(1\ 2\ 6\ 4)$ in $\mathrm S_6$ is the permutation $$\begin{pmatrix} 1 &2 &3 &4 &5 &6\\ \downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow\\ 2 &6 &3 &1&5 &4 \end{pmatrix}$$ Now, for any $\sigma\in\mathrm S_n$, do the following.

1. Look at the list $1$, $\sigma(1)$, $\sigma^2(1)$, $\dots$. This has to start repeating at some point, suppose $\sigma^k(1)=1$. Then, let $\sigma_1=(1\ \sigma(1)\ \sigma^2(1)\ \cdots\ \sigma^k(1))$.
2. If $k=n$, then $\sigma=\sigma_1$ (why?). Otherwise, there is at least one element which is not among the $\sigma^i(1)$s. Let that be $b$. Then, make the list $b$, $\sigma(b)$, $\sigma^2(b)$, $\dots$. Say it repeats at $\ell$, i.e., $\sigma^\ell(b)=b$. Then, consider $\sigma_2=(b\ \sigma(b)\ \sigma^2(b)\ \cdots\ \sigma^\ell(b))$. (Check that none of these can be same as what we had got before.)
3. If $k+\ell=n$, then $\sigma=\sigma_1\circ\sigma_2$ (why?). Else repeat.

This will terminate and you will get $\sigma=\sigma_1\circ\sigma_2\circ\cdots\circ\sigma_r$ (why?).

As an example, take the $\sigma$ in question. The first list is $1$, $4$, $3$, $1$, $\dots$ so $\sigma_1=(1\ 4\ 3)$. Then, $b=2$ and $\sigma_2=(2\ 5)$. Check that $\sigma$ is the same permutation as the composition of $(1\ 4\ 3)$ and $(2\ 5)$. You can omit the $\circ$s and simply write $$\sigma=(1\ 4\ 3)(2\ 5)$$

---

For the order of a permutation, your method is correct since you found that $\sigma^6=1$ but $\sigma^k\neq1$ for $1\leq k\le 5$.

Here is a sequence of exercises to help you find an order of a permutation from its cycle decomposition (the one we derived before).

1. Show that order of $(a_1\ a_2\ \cdots\ a_k)$ is $k$.
2. Show that if $\sigma=\sigma_1\sigma_2\dots\sigma_r$ is the cycle decomposition, then $$\sigma^2=\sigma_1^2\sigma_2^2\dots\sigma_r^2$$
3. Generalise 2 and use it to derive a formula for the order of $\sigma$ in terms of lengths of the cycles.

Hope this helps. :)