Hello,

I am stuck on part three of this question.

i) $\displaystyle \sigma $ = (1 6) (1 4) (1 2) (3 7) (3 5) (8 9) = even permutation

ii) $\displaystyle \sigma^{2} $ = 1 2 3 4 5 6 7 8 9

4 6 7 1 3 2 5 8 9

III) how do I find T??

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- Jan 27th 2014, 01:04 PMTweetypermutation help ,
Hello,

I am stuck on part three of this question.

i) $\displaystyle \sigma $ = (1 6) (1 4) (1 2) (3 7) (3 5) (8 9) = even permutation

ii) $\displaystyle \sigma^{2} $ = 1 2 3 4 5 6 7 8 9

4 6 7 1 3 2 5 8 9

III) how do I find T?? - Jan 27th 2014, 01:14 PMSlipEternalRe: permutation help ,
You do it digit by digit.

You want: $\displaystyle \tau \sigma(1) = 2$. You have $\displaystyle \tau \sigma(1) = \tau(\sigma(1)) = \tau(2)$, so $\displaystyle \tau(2) = 2$.

You want: $\displaystyle \tau \sigma(2) = 1$. You have $\displaystyle \tau \sigma(2) = \tau(\sigma(2)) = \tau(4)$, so $\displaystyle \tau(4) = 1$.

Etc. - Jan 27th 2014, 02:14 PMTweetyRe: permutation help ,
oh okay, so the rest just stays the smame?

- Jan 27th 2014, 02:21 PMSlipEternalRe: permutation help ,
Turning two-line notation into three-line notation:

$\displaystyle \tau \sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}$

Essentially, the first two lines show where $\displaystyle \sigma$ sends numbers while the second two lines show where $\displaystyle \tau$ sends numbers. So, the first and last line show where the composition $\displaystyle \tau \sigma$ sends numbers. To write $\displaystyle \tau$, just write the last two lines, but order the columns so that the top row is in order:

$\displaystyle \tau = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 2 & 7 & 1 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$ - Jan 27th 2014, 02:49 PMPlatoRe: permutation help ,
- Jan 27th 2014, 03:23 PMSlipEternalRe: permutation help ,
In case it is not obvious to the OP, $\displaystyle \sigma^{-1}$ is extremely easy to calculate. You are given that $\displaystyle \sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8\end{pmatrix}$, so its inverse is just flipping the two rows:

$\displaystyle \sigma^{-1} = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$

So, now, the product (composition) of the two permutations: $\displaystyle \tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$ is easy to calculate (and gives the same answer as above). Generally, as you continue in mathematics, Plato's solution is going to be easier to apply. - Jan 27th 2014, 03:35 PMTweetyRe: permutation help ,
- Jan 27th 2014, 03:39 PMSlipEternalRe: permutation help ,
$\displaystyle \tau\sigma = (1 2)$, so multiplying both sides by $\displaystyle \sigma^{-1}$ on the right, you get $\displaystyle \tau \sigma \sigma^{-1} = (1 2) \sigma^{-1}$. On the LHS, $\displaystyle \sigma \sigma^{-1}$ is the identity permutation, so the LHS simplifies to $\displaystyle \tau$. So, it is not that $\displaystyle \sigma \sigma^{-1} = \tau$, but that you are multiplying $\displaystyle \tau$ by the identity permutation, so you are not changing it. If you send 1 to 2, then send 2 back to 1, it is the same as if you never sent it anywhere at all. That is how multiplying a permutation by its inverse gives the identity permutation.

- Jan 27th 2014, 03:40 PMTweetyRe: permutation help ,
thanks I understand now!