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I need to show that:

  1. $S=\left\{(12),(13),...,(1n)\right\}$ generates $S_n$
  2. $S=\left\{(12),(123\cdots n)\right\}$ generates $S_n$

How do I show that each one of them generates $S_n$?

Thank you!

Mikasa
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CS1
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1 Answers1

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Can you prove that every element of $S_n$ is equal to a product of transpositions? If so, you just need to show that each of those generating sets contains all transpositions.

Edit: To make this more direct, this task becomes straight forward once you understand how conjugation works in $S_n$.

JMag
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  • This is the first one, how I solved the second? – CS1 Nov 25 '13 at 20:54
  • Try conjugating $(12)$ by $(123...n)$. The result is

    $(123...n)(12)(123...n)^{-1} = (23)$. Depending on how you multiply (from left to right or right to left) this equation may be incorrect. However, this is the general idea.

     – JMag Nov 25 '13 at 20:56
  • Read lemma 2 on page 2 of http://www.math.wm.edu/~vinroot/415conj.pdf. This is the general formula for conjugation in $S_n$. – JMag Nov 25 '13 at 21:00
  • Yes, but this gives me only the $(ab)$ cycles, no? – CS1 Nov 25 '13 at 21:10
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    Well, it will actually give ${(12),(23),...(n-1,n),(n,1)}$. From this set you can generate all other transpositions. For example $$(23)(12)(23)=(13)$$ and $$(34)(13)(34)=(14).$$ – JMag Nov 25 '13 at 21:30