# 4th symmetric group, order

• Jun 11th 2009, 10:39 PM
Maccaman
4th symmetric group, order
I have the following revision question:

Recall that $S_4$ (the fourth symmetric group) consists of the set of bijections $f: \{ 1,2,3,4 \} \rightarrow \{ 1,2,3,4 \}.$ The group operation is composition of functions. Let
$p = \begin{pmatrix} 1&2&3&4 \\ 4&2&1&3 \end{pmatrix}$

Find the order of $p$ and write down all the elements of the cyclic subgroup $

$

.

I think that to find the order of $p$ I need to calculate $p^{something} = 1.$ i.e. $p^{something} = \begin{pmatrix} 1&2&3&4 \\ 1&2&3&4 \end{pmatrix}$ and then the order is the $something$

Then I think that if, for example, the order was 2 then $

= \{ p,p^1 \}....$

but Im not sure.

Anyway my main problem is calculating $p^2, p^3, p^4, ect$. Can anyone please show me how this is done? Thanks
• Jun 12th 2009, 01:30 AM
flyingsquirrel
Quote:

Originally Posted by Maccaman
Then I think that if, for example, the order was 2 then $

= \{ p,p^1 \}....$

but Im not sure.

Don't forget the identity element: $\langle p \rangle =\{1,p,p^2,\ldots,p^{\text{something}-1}\}$ (and "something" is the least positive integer such that $p^\text{something}=1$).
Quote:

Originally Posted by Maccaman
Anyway my main problem is calculating $p^2, p^3, p^4, ect$. Can anyone please show me how this is done? Thanks

We have $\begin{pmatrix} {\color{blue}1}&{\color{green}2}&3&{\color{red}4} \\ {\color{blue}4}&{\color{green}2}&1&{\color{red}3} \end{pmatrix}$. To compute $p^2$ we need to compute $p^2(1),\ p^2(2),\ p^2(3)$ and $p^2(4)$.

• $p^2(1)=p( p(1)) = p(4)=3$ because ${\color{blue}p(1)=4}$ and ${\color{red}p(4)=3}$.
• $p^2(2)=p( p(2)) = p(2)=2$ because ${\color{green}p(2)=2}$.
• $p^2(3)=p( p(3)) = p(1)=4$ because ${p(3)=1}$ and ${\color{blue}p(1)=4}$.
• $p^2(4)=p( p(4)) = p(3)=1$ because ${\color{red}p(4)=4}$ and ${p(3)=1}$.

hence
$p^2=\begin{pmatrix} 1&2&3&4 \\ 3&2&4&1 \end{pmatrix}$

Can you take it from here ?