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I have a question on the following exercise presented in Dummit and Foote, $\S 3.5$. The exercise is stated as follows:

Problem:

Show that if $p$ is prime, $S_p = \langle \sigma, \tau \rangle$ where $\sigma$ is any transposition and $\tau$ is any $p$-cycle.

My question is how can any $p$-cycle be an element of $S_p$? Since $\sigma$ is a generator of $S_p$ we must have that $\sigma \in S_p$. But consider $(4\ 8\ 9\ 10\ 11)$. This is a $5$-cycle but it is not a member of $S_p$. What am I missing here? It would seem this is a counter example showing the theorem to be false, which I am skeptical of!

Thanks

balddraz
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H_1317
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  • I am thinking it is just sloppy wording, as any transposition would also cause similar errors in that (99 100) is not a member of S$_5$ also. I think the author just means p-cycles and transpositions with integers less than or equal to p. Does anyone agree? – H_1317 Jul 28 '19 at 14:31
  • They mean any p-cycle in $S_p$. You are correct. –  Jul 28 '19 at 14:34
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    Usually we represent $S_n$ as the permutations of numbers $1$ to $n$, in your example $(4 \ 8 \ 9 \ 10\ 11)$ is not an element of $S_5$ as we usually represent it, although you could just decide to represent $S_5$ as the set of permutations of ${4, 8, 9, 10, 11}$. – DottorMaelstrom Jul 28 '19 at 14:36
  • $S_n$ is generated by the cycle $(1, 2, 3 \ldots n)$ and the transposition $(1 ,2)$ (because it contains all the transpositions $(1, m)$ and $(k, m)$). If we have a cycle and a transposition we can choose the numbering such that they are $(1, 2, 3 \ldots n)$ and $(1, m)$ and they generate $S_n$ if $\gcd(m-1,n)=1$, which is always satisfied if $n$ is prime. – reuns Jul 28 '19 at 15:44

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