Prove or disprove: $S_{10} = \langle (1,3),(1,2, ... ,10) \rangle$
I know that $S_{10}=\langle (1,2) , (1,2,...,10) \rangle$. I tried to use this fact to prove the above but failed. It made me think it is not true, but I couldn't find an explanation. Any thoughts?
Look at the action of your group on $\{1,2,\ldots,10\}^2=\{[1,1],[1,2],[1,3],\ldots[10,10]\}$. (I'm using brackets for the ordered pairs, since parentheses indicate cycles in this topic.) The action is just coordinate-wise.
Of your two elements, $(1,2,\ldots,10)$ takes $[\text{odd},\text{odd}]$ to $[\text{even},\text{even}]$ and vice versa. The other one, $(1,3)$ takes any $[\text{odd},\text{odd}]$ to some possibly different $[\text{odd},\text{odd}]$, and leaves $[\text{even},\text{even}]$ unmoved.
Therefore this group has no elements that could take $[\text{odd},\text{odd}]$ to $[\text{odd},\text{even}]$. In particular, it cannot take $[3,1]$ to $[3,2]$. But if the group were the full $S_{10}$, it would have the transposition $(1,2)$, which does take $[3,1]$ to $[3,2]$. So the group is missing $(1,2)$ among many other elements of $S_{10}$.