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I am having trouble finding good resources to understand composition of permutation, I was wondering how can you go about multiplying (composition) of two permutations and how do they return the just a different permutation of the set?

In the picture, in the answer sheet to the right, where does the third line in the first 'set' come from?

https://i.sstatic.net/RARyI.png

For the inverse of a permutation: since the permutation is cycle it means that any random arrangement of the set can be passed of as an inverse? It says interchanging the rows finds inverse but in the second line its random

And in the third question, does id refer to identity i.e. 1?

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  • $\begingroup$ Do you understand that a permutation is simply a particular kind of function from a set to itself? For instance, the permutation $\sigma$ in Problem $5$ is the function from the set $\{1,2,3,4,5,6,7,8\}$ to itself such that $\sigma(1)=3$, $\sigma(2)=2$, $\sigma(3)=6$, and so on. Multiplying permutations is simply composing them as functions: applying one function and then the other. $\endgroup$
    – Brian M. Scott
    Commented Feb 15, 2013 at 21:43
  • $\begingroup$ @Brian: Yes, I was thinking I should probably really have started my answer with an explanation of what a permutation actually is, rather than just going through the questions listed here. $\endgroup$
    – Tara B
    Commented Feb 16, 2013 at 9:33

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Any basic group theory or abstract algebra textbook should be good for learning about permutations. I don't know what people tend to use these days. In my abstract algebra course we used Contemporary Abstract Algebra by Gallian.

The third line in 5a) on the answer sheet comes from applying the permutation $\sigma$ to the numbers in the second row.

"For the inverse of a permutation: since the permutation is cycle it means that any random arrangement of the set can be passed of as an inverse? It says interchanging the rows finds inverse but in the second line its random"

This part of your question is a bit confused. Firstly, not every permutation is a cycle, but I'm not sure whether you were saying that (I don't see anything in the linked image about inverses of permutations, so I'm not sure what 'the permutation' would be referring to).
Every permutation has a unique inverse, so the answer to your question is definitely no!
I don't know what you mean by 'but in the second line it's random'.

Here is an example of how to find the inverse of a permutation by interchanging the rows.

If $\sigma = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\2 & 4 & 1 & 3 \end{array}\right),$ then $\sigma^{-1} = \left(\begin{array}{cccc} 2 & 4 & 1 & 3\\1 & 2 & 3 & 4 \end{array}\right) = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\3 & 1 & 4 & 2 \end{array}\right)$.

So we first interchange the rows, and then rearrange the columns so that the numbers in the top row are in order, as we are accustomed to seeing them. The reason this works is that if $\sigma$ takes $x$ to $y$, then $\sigma^{-1}$ needs to take $y$ back to $x$ in order to invert the permutation. (Check by calculating $\sigma\sigma^{-1}$ and making sure you get the identity!)

Yes, 'id' does refer to the identity, i.e. the permutation that fixes every point.

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  • $\begingroup$ The third line comes from applying $\tau$. $\endgroup$
    – Git Gud
    Commented Feb 15, 2013 at 18:25
  • $\begingroup$ @GitGud: Not in 5a)! $\endgroup$
    – Tara B
    Commented Feb 15, 2013 at 18:34
  • $\begingroup$ The third line in $\tau \circ \sigma$? If so, I stand by my statement. $\endgroup$
    – Git Gud
    Commented Feb 15, 2013 at 18:35
  • $\begingroup$ No, in the first bit right at the beginning of 5a) (so in $\sigma\circ \tau$). I presumed that was what the OP meant by "the first 'set' ". $\endgroup$
    – Tara B
    Commented Feb 15, 2013 at 18:43
  • $\begingroup$ I missed that. ${}{}{}$ $\endgroup$
    – Git Gud
    Commented Feb 15, 2013 at 18:45

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