Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.

Question

$\sigma = (13)(254), \sigma^2=(13)(254)(13)(254) = (13)(13)(254)(254) = (425) $

Or, in two row format, get: $$ \sigma = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} $$

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

Due to disjoint cycles, they can commute as no affect on map produced.

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

My answer is:

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

But, the answer is different: $(245)$?

@ChristopheLeuridan then second too is, as just two row format. But, how? — jiten, Jun 10 '22 at 17:52
How is your answer different from $(2~4~5)$? In your answer $2 \mapsto 4$, $4\mapsto 5$ and $5 \mapsto 2$ which is exactly $(2~4~5)$. — Randy Marsh, Jun 10 '22 at 17:57

score 2 · Accepted Answer · edited Jun 10 '22 at 17:57

2

Here

$$\begin{align} \sigma^2&=\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\cdot\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\\ &=\begin{pmatrix}1&2&3&4&5\\1&4&3&5&2\end{pmatrix}\\ &=(245). \end{align}$$

edited Jun 10 '22 at 17:57

Shaun

44,997

answered Jun 10 '22 at 17:53

Sourav Ghosh

12,997

score 2 · Answer 2 · answered Jun 10 '22 at 18:09

We have

$$\begin{align} \sigma^2&=(13)(254)(13)(254)\\ &=(13)(13)(254)(254)\\ &=(254)^2\\ &=(245) \end{align}$$

because

$$\begin{align} 1&\xrightarrow{(254)}1\xrightarrow{(254)}1,\\ 2&\xrightarrow{(254)}5\xrightarrow{(254)}4,\\ 3&\xrightarrow{(254)}3\xrightarrow{(254)}3,\\ 4&\xrightarrow{(254)}2\xrightarrow{(254)}5,\\ 5&\xrightarrow{(254)}4\xrightarrow{(254)}2.\\ \end{align}$$

In row format,

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

which gives $(245)$ again.

Very happy, as was unable to understand your answer at: https://math.stackexchange.com/a/3787695/513178. — jiten, Jun 10 '22 at 18:16

Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.

2 Answers2