I was hoping someone could help me understand the following proof from:(adsbygoogle = window.adsbygoogle || []).push({});

http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf

The problem I'm having is the line ``Any p-cycle can be written as ##(1\;2\;\cdots\;p)## by relabeling the objects being permuted by applying an overall conjugation on ##S_p##''. Corollary 2.10. For a prime number ##p##, ##S_p## is generated by any transposition and any p-cycle. Proof. Any p-cycle can be written as (12...p) by relabeling the objects being permuted (that means by applying an overall conjugation on ##S_p##), so to show any transposition and any p-cycle generate ##S_p## it suffices to show any transposition and the standard p-cycle (12...p) generate ##S_p##

An example helps me to understand:

Suppose I have ##R=(3\;5),Q=(4\;1\;2\;5\;3)\in S_5##. Now, I take this to mean if I conjugate the entire group, ##\sigma S_5 \sigma^{-1}## such that ##\sigma Q \sigma^{-1}=(1\;2\;3\;4\;5)##, then I've effectively ``relabeled'' ##Q## with ##(1\;2\;3\;4\;5)##. That's easy to accomplish since in general for ##\rho=(a\;b\;c\;d\;e)##, ##\sigma\rho\sigma^{-1}=(\sigma(a)\;\sigma(b)\;\sigma(c)\;\sigma(d)\;\sigma(e))##. Then let ##\sigma=(1\;2\;3\;5\;4)## and then we have ##(1\;2\;3\;5\;4)(4\;1\;2\;5\;3)(4\;5\;3\;2\;1)=(1\;2\;3\;4\;5)##. Therefore I assume this all means:

##

\big<(3\;5),(4\;1\;2\;5\;3)\big>=\big<(h\;k),(1\;2\;3\;4\;5)\big>

##

since conjugation is an automorphism (it's a bijection) and where ##(h\;k)## is the conjugation of ##(3\;5)## and just accepting for the moment if I can get it to that form, it generates the group via the second part of the theorem which I will accept for now.

Am I interpreting this correctly?

Thanks for reading,

Jack

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# Showing any transposition and p-cycle generate S_p

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