# How to show that two groups makes $S_n$

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!

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"Generates" ${}{}$ – Casteels Nov 25 '13 at 20:42
I fix it, thank you! – Yoar Nov 25 '13 at 20:54
See also this great post of missing Arturo. – Babak S. Nov 26 '13 at 7:14

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 $S_n$.
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 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? – Yoar Nov 25 '13 at 21:10
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