Is $S_5$ generated by (1 3) and (1 2 3 4 5)?

Question

I proved that $S_4$ is not generated by $(1 \ 3)$ and $(1 \ 2 \ 3 \ 4)$, using that the partition $A = \{ \{ 1, 3 \}, \{ 2, 4 \} \}$ is invariant under the action of the elements of $\left \langle (1 \ 3), (1 \ 2 \ 3 \ 4) \right \rangle$, but if we take a permutation $\sigma \notin \left \langle (1 \ 3), (1 \ 2 \ 3 \ 4) \right \rangle$, for example $\sigma = (1 \ 2)$, then $\sigma A \ne A$. I tried to use the same reasoning for $S_5$, taking a partition $P = \{ \{1, 3 \}, \{ 2, 4, 5 \} \}$ and I got that $P$ is invariant under the action of $(1 \ 3)$, but isn't invariant under the action of $(1 \ 2 \ 3 \ 4 \ 5)$, that is, $(1 \ 2 \ 3 \ 4 \ 5)P \ne P$. Any tips to help me?

Answer 1

Start by recalling that, for $\sigma,\tau\in S_n$, $$\tau=(a_1,a_2,...,a_k) \implies \sigma\tau\sigma^{-1} = (\sigma(a_1),\sigma(a_2),...,\sigma(a_k))$$ For example, setting $\sigma=(1\ 2\ 3\ 4\ 5)$ and $\tau=(1\ 3)$, we have $$\tau=(1\ 3),$$ $$\sigma\tau\sigma^{-1}=(2\ 4),$$ $$\sigma^2\tau\sigma^{-2}=(3\ 5),$$ $$\sigma^3\tau\sigma^{-3}=(4\ 1),$$ $$\sigma^4\tau\sigma^{-4}=(5\ 2)$$ So, within $\langle \sigma,\tau \rangle$ we are able to transpose all elements at a distance of $2$ in either direction. This means that we can also transpose all elements at a distance of $2k, k\in\mathbb{Z}$, since of course $$s_1=(i,\ i+2), s_2=(i+2,\ i+4)\implies s_1s_2s_1 =(i,\ i+4)$$ But in $\mathbb{Z_5}$, a distance of 4 in one direction is equal to a distance of 1 in the other, so we can now also swap any adjacent elements. Therefore the set of all adjacent transpositions is inside $\langle \sigma,\tau \rangle$, and so $S_5\subset \langle \sigma,\tau \rangle$, and we are done.

Note: The reason this didn't work in $S_4$ is that moving in a distance of $2k$ in either direction doesn't allow you to reach every element the same way it does in $S_5$.