Showing any transposition and p-cycle generate S_p

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The discussion revolves around understanding a proof that any transposition and any p-cycle generate the symmetric group S_p for a prime number p. The key point is that any p-cycle can be expressed as a standard form (1 2 ... p) through conjugation, which effectively relabels the elements being permuted. An example using specific permutations illustrates how conjugation works, confirming that the generated group remains the same. The participant confirms their interpretation aligns with the theorem, emphasizing the importance of expressing any group element in terms of a chosen transposition and the standard p-cycle. Overall, the discussion clarifies the relationship between transpositions, p-cycles, and the generation of symmetric groups.
jackmell
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I was hoping someone could help me understand the following proof from:

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

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##''.

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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Yes.

The second part of the theorem is proven as Theorem 2.8 of what you linked.

Given that, you just need to write out the formulas by which an arbitrary element g of the group can be expressed in terms of a transposition T and a p-cycle C by first using Theorem 2.8 to express the conjugate of g in terms of another transposition (why not choose (1 2) WLOG) and the standard p-cycle.
 
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