How do i prove how $S_5$ is generated by a two cycle and a five cycle? 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 $(12345)$? 
 A: 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.
A: 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 $(1, 2, \dots, n)$.
A: 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 $(1 2 3 4 5)=(1 5)(1 4)(1 3)(1 2)$
