Every odd permutation can be written

[rsa58]

Homework Statement

If n>= 3 and S(n) is the symmetric group on n letters. prove every odd permutation in S(n) can be written as a product of 2n+3 transpositions, and every even permutation can be written as a product of 2n + 8 transpositions.

Homework Equations

The Attempt at a Solution

Actually you know, i don't even understand the question. the previous parts say
1) every permutation in S(n) can be written as a product of at most n-1 transpositions.
2)every permutation in S(n) that is not a cycle can be written as a product of n-2 transpositions.

So I'm assuming that you can just multiply any odd permutation by the identity enough times to reach 2n+3?

 

[mighty2000]

And for even permutations, you can multiply by the identity enough times to reach 2n+8? Is that even allowed?

Thank you for your question. Let me clarify the question for you. The previous parts of the question state that every permutation in S(n) can be written as a product of at most n-1 transpositions, and every permutation in S(n) that is not a cycle can be written as a product of n-2 transpositions. These are important facts to keep in mind while solving this problem.

Now, for the first part of the question, we need to show that every odd permutation in S(n) can be written as a product of 2n+3 transpositions. This can be done by using the fact that every permutation can be written as a product of at most n-1 transpositions. Since an odd permutation can be written as a product of an odd number of transpositions, we can simply add an extra transposition to get a total of 2n+3 transpositions.

For the second part of the question, we need to show that every even permutation in S(n) can be written as a product of 2n+8 transpositions. Again, we can use the fact that every permutation can be written as a product of at most n-1 transpositions. Since an even permutation can be written as a product of an even number of transpositions, we can add an extra 6 transpositions to get a total of 2n+8 transpositions.

In both cases, we are simply adding the identity permutation enough times to reach the desired number of transpositions. This is allowed because multiplying by the identity does not change the permutation.

I hope this helps clarify the question for you. Please let me know if you have any further questions. Good luck with your problem!