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How do I prove that $S_5$ (the permutation group on five letters) can be generated by a two-cycle $(12)$ and a five cycle $(12345)$?

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  • $\begingroup$ It's enough to show that you can produce every transposition. That is, find a way to generate each of $$ (13),(14),(15),(23),(24),(25),(34),(35),(45) $$ $\endgroup$ – Omnomnomnom Apr 28 '14 at 3:47
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    $\begingroup$ @user146285 regarding that theorem for when $n$ is prime: $S_n$ is generated by $(12)$ and $(123\cdots n)$ for all $n$; the proof is very similar to what you see in the answers below. In the prime case, it is true that any two cycle together with any $n$-cycle generate $S_n$, because you can do a substitution to convert the $n$-cycle into one which, together with the given 2-cycle, "looks like" the above pair of generators. This can't necessarily be done if $n$ is composite, for example, $(1234)$ and $(13)$ do not generate $S_4$. $\endgroup$ – Dustan Levenstein Apr 28 '14 at 3:58
  • $\begingroup$ (My comment is in response to a deleted comment.) $\endgroup$ – Dustan Levenstein Apr 28 '14 at 3:59
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Define $s = (12345)$ and $t = (12)$. Note that $$ s\, t\, s^{-1} = (23) $$ Similarly, $$ s^2 t s^{-2} = (34), \quad s^3 t s^{-3} = (45), \quad s^4 t s^{-4} = (51) $$ From there, it's not too hard to get the rest of them. For example, $$ (13) = (23)(12)(23) $$ In this manner, we show that every transposition can be generated by $s$ and $t$. Thus, all of $S_5$ is generated by $s$ and $t$.

Lemma 2 over here should help you understand what's going on.

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First, verfiy the following equalities: \begin{align*} (12345)(12)(12345)^{-1} &= (23)\\ (12345)^2(12)(12345)^{-2} &= (34)\\ (12345)^3(12)(12345)^{-3} &= (45). \end{align*} You can then use these three equalities to generate all transpositions, and hence all of $S_5$.

It actually turns out that this result can be generalized. Any symmetric group $S_n$ is generated by $(12)$ and $(1, 2, \dots, n)$.

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Proof that $S_{n}$ can be written as a product of transpositions:

Let $(1 2 3 ... n)\in S_{n}$ be a cycle of length n.

Now, $(1 2 3 ... n-1 n)=(1 n)(1 (n-1))...(1 4)(1 3)(1 2)$

For example, $(7 6 3)$ can be written as $(7 3)(7 6)$

$(1 2 3 4 5)\in S_{5}$. Then $(1 2 3 4 5)=(1 5)(1 4)(1 3)(1 2)$

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