Suppose $G$ is a transitive subgroup of $S_n$ such that it there exist $\sigma, \tau \in G$ such that $\sigma$ is an $(n-1)$-cycle and $\tau$ is a transposition. Prove that $G = S_n$.
I just don't understand how to mathematically use the transitive nature of the subgroup.
Best Answer
Take your subgroup $G$, up to the study of a conjugate $G$ you can assume that the $n-1$-cycle of $G$ is $c=(2,...,n)$. Now if $\tau$ is a transposition in $G$ then $\tau=(i,j)$ with $i\neq j$. Take $\sigma_i\in G$ such that $\sigma_i(i)=1$ (this is where I use the transitivity).
Then I claim that $\sigma_i\tau\sigma_i^{-1}=(1,k)$ where $k\geq 2$. Now you have $\tau_0=(1,k)$ where $k\geq 2$ and $c=(2,...,n)$ in $G$, I claim that :
$$\{c^s\tau_0c^{-s}|s\in\mathbb{N}\}=\{(1,2),...,(1,n)\} $$
This shows that $G$ will contain $(1,2),...,(1,n)$ (because $c^s\tau_0c^{-s}\in G$, $c$ and $\tau_0$ are in $G$). Now it is easy to see that $\{(1,2),...,(1,n)\}$ is a generating set of $S_n$, hence $G=S_n$.