Algebraic structures

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Whitespy

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First of all I am new in this forum im 20 years old and I am studying mathematics in an university in Greece.
I do not know if i post it at right category but I think it's okey.

I have problem with these 2 exercises and I don't know/understand the steps to find this answers.

1)Find the solution: τ=(1,2,5,7)(3,4,6,7)(1,4,8,9,3)(4,7,2)
answer: (1,7)(2,8,9,4,3,6)(5) => (2,8,9,4,3,6)(1,7)

2)Find the solution:σ=(1,3,5)(2,7)(11,1,4)(5,6,2,8,9)(7,3)(4,5,2,1)
answer: (1,11,3,2,4,6,7,5,8,9)(10) =>(1,11,3,2,4,6,7,5,8,9)

τ,σ are permutations if it was not perceived.

Thank you for your time.
 
What is meant by "solution"? What were you told to do? If you are to write each permutation as a product of disjoint cycles, I don't get the answer you state for the first (the only one I've tried). WolframAlpha gives the same result I got by hand.

That could be because you are defining things differently than I (or they) do. Please state how your notation is defined, especially the order in which you compose cycles.

Once we are sure of the question and your definitions, we can talk about how to do it. But I should also ask what you have learned or tried. Were you given any examples? Did you make an attempt you can show us?
 
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Look for example at the first τ have (1,2,5,7)(1,4,8,9,3) ,1->2 at the first permutation but at the second permutation 1->4 .Every element must be only one time.

12412
At the permutation rules why 1->4 and not 1->2 like the 1st permutation?
 
Dr.Peterson said:
What is meant by "solution"? What were you told to do? If you are to write each permutation as a product of disjoint cycles, I don't get the answer you state for the first (the only one I've tried). WolframAlpha gives the same result I got by hand.
That could be because you are defining things differently than I (or they) do. Please state how your notation is defined, especially the order in which you compose cycles. Once we are sure of the question and your definitions, we can talk about how to do it. But I should also ask what you have learned or tried. Were you given any examples? Did you make an attempt you can show us?
Click to expand...
I had almost the identical reaction as Prof Peterson's. Except I did not try WolframAlpha. Thanks for that because I had "multiplied" \(\displaystyle \tau\) out & it is great to know my result agrees WA's. So please do, Whitespy, help us out here.
 
Yes my both exp are wrong but by the way I don't know how from the 1st arrow we go in the result.
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Obviously I don't have understand something.
 
You have to do one permutation after another; I'm assuming left to right like WA. You don't look at what they all do to any given element, but at what each does to what the given element has already been taken to.

Starting with 1, the first permutation takes that to 2; the next two ignore 2; and the last takes 2 to 4. So, together, they take 1 to 4, as indicated in the two-line notation.

Similarly, (1 2 5 7) takes 2->5; then nothing else moves is, so we have the second column.

Does your textbook give no explanation or examples?

By the way, here is a nice explanation of a couple different ways to approach it, starting slowly.
 
Here is an example from Herstein: \(\displaystyle (12)(123)(45)(568)(179)\)
\(\displaystyle \left( {\begin{array}{*{20}{c}}1&2&3&4&5&6&7&8&9\\ - & - & - & - & - & - & - & - & - \end{array}} \right)\)
The idea is to fill in the blanks. Be careful. Not all authors agree on the left-to-right or right-to-left convention.
I will do the first four.
\(\displaystyle 1\to 2,~2\to 3,~3\to 3,~3\to 3,~3\to 3\) so \(\displaystyle 1\to 3\)
\(\displaystyle 2\to 1, 1\to 2\to 2, 2\to 2, 2\to 2, 2\to 2, 2\to 2\) so \(\displaystyle 2\to 2\)
\(\displaystyle 3\to 3, 3\to 1, 1\to 1, 1\to 1, 1\to 7,\) so \(\displaystyle 3\to 7\)
\(\displaystyle 4\to 4, 4\to 4, 4\to 5, 5\to 6, 6\to 6,\) so \(\displaystyle 4\to 6\)

Hence: \(\displaystyle \left( {\begin{array}{*{20}{c}}1&2&3&4&5&6&7&8&9\\ 3 & 2 & 7 & 6 & - & - & - & - & - \end{array}} \right)\)
Now you try to finish and post what you get. If in doubt check HERE

 
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