# Transitive subgroup of symmetric group $S_n$ containing an $(n-1)$-cycle and a transposition

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.

• $\sigma$ is an $n$-cycle I think. – Yilong Zhang Jun 23 '15 at 9:26
• Sorry, I missed "transitive". – Yilong Zhang Jun 23 '15 at 9:28
• @YilongZhang No, it is $n-1$. Just out of curiosity, can you prove the statement if it is $n$? – MathManiac Jun 23 '15 at 9:28
• @YilongZhang, when $n=4$ the $2$-Sylows of $S_4$ contain a transposition and a $4$-cycle, but they are not the whole group. Think of $\langle(1,2,3,4),(1,3)\rangle$ (However the property is true if $n$ is prime). – Clément Guérin Jun 23 '15 at 9:30
• @ClémentGuérin Thanks a lot for your comment. I was just under the impression that every pair of $n$-cycle and transposition can generate $S_n$. – Yilong Zhang Jun 23 '15 at 9:51

## 1 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 $G=S_n$.