permutation help ,

• Jan 27th 2014, 01:04 PM
Tweety
permutation help ,
Hello,

I am stuck on part three of this question.

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

ii) $\sigma^{2}$ = 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 PM
SlipEternal
Re: permutation help ,
You do it digit by digit.

You want: $\tau \sigma(1) = 2$. You have $\tau \sigma(1) = \tau(\sigma(1)) = \tau(2)$, so $\tau(2) = 2$.
You want: $\tau \sigma(2) = 1$. You have $\tau \sigma(2) = \tau(\sigma(2)) = \tau(4)$, so $\tau(4) = 1$.
Etc.
• Jan 27th 2014, 02:14 PM
Tweety
Re: permutation help ,
oh okay, so the rest just stays the smame?
• Jan 27th 2014, 02:21 PM
SlipEternal
Re: permutation help ,
Turning two-line notation into three-line notation:

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

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

$\tau = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 2 & 7 & 1 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$
• Jan 27th 2014, 02:49 PM
Plato
Re: permutation help ,
Quote:

Originally Posted by Tweety
III) how do I find T??

Here is how I would have done it.

First find $\sigma^{-1}$ then $\tau = \left( {\begin{array}{*{20}{c}}1&2&3&4&5&6&7&8&9\\2&1&3&4 &5&6&7&8&9\end{array}} \right) \circ {\sigma ^{ - 1}}$

That gives the same as above.
• Jan 27th 2014, 03:23 PM
SlipEternal
Re: permutation help ,
Quote:

Originally Posted by Plato
Here is how I would have done it.

First find $\sigma^{-1}$ then $\tau = \left( {\begin{array}{*{20}{c}}1&2&3&4&5&6&7&8&9\\2&1&3&4 &5&6&7&8&9\end{array}} \right) \circ {\sigma ^{ - 1}}$

That gives the same as above.

In case it is not obvious to the OP, $\sigma^{-1}$ is extremely easy to calculate. You are given that $\sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8\end{pmatrix}$, so its inverse is just flipping the two rows:

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

So, now, the product (composition) of the two permutations: $\tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$ 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 PM
Tweety
Re: permutation help ,
Quote:

Originally Posted by SlipEternal
In case it is not obvious to the OP, $\sigma^{-1}$ is extremely easy to calculate. You are given that $\sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8\end{pmatrix}$, so its inverse is just flipping the two rows:

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

So, now, the product (composition) of the two permutations: $\tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$ 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.

Thank you,

the other solution is easier to understand but now I dont understand why $\sigma \sigma^{-1} = \tau$?
• Jan 27th 2014, 03:39 PM
SlipEternal
Re: permutation help ,
$\tau\sigma = (1 2)$, so multiplying both sides by $\sigma^{-1}$ on the right, you get $\tau \sigma \sigma^{-1} = (1 2) \sigma^{-1}$. On the LHS, $\sigma \sigma^{-1}$ is the identity permutation, so the LHS simplifies to $\tau$. So, it is not that $\sigma \sigma^{-1} = \tau$, but that you are multiplying $\tau$ 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 PM
Tweety
Re: permutation help ,
thanks I understand now!