# 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 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. Jun 23, 2015 at 9:26
• @YilongZhang No, it is $n-1$. Just out of curiosity, can you prove the statement if it is $n$? Jun 23, 2015 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). Jun 23, 2015 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 $S_n$. Jun 23, 2015 at 9:51
• math.stackexchange.com/questions/96298/… Mar 26, 2017 at 21:51

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 $$G=S_n$$.