I have a doubt about the simplicity of the alternating group $A_n$ in Dummit&Foote's Abstract Algebra on page150 :
It's to prove that the alternating group $A_n=G$ is simple for $n\geq5$ . Let $H\unlhd G$ and the author shows that:
if $\tau\ne1$ is an element of $H$ then $\tau(i)\ne i$ for all $i\in\{1,2,\cdots,n\}$ , i.e., no nonidentity element of $H$ fixes any element of $\{1,2,\cdots,n\}$ .
It also implies that there exists no 1-cycles in the cycle decomposition of each nonidentity element in $H$ . But I am confused with the following assertion which is also given by the author:
It follows that if $\tau_1,\tau_2$ are elements of $H$ with $\tau_1(i)=\tau_2(i)$ for some $i$ , then $\tau_1=\tau_2$ , since then $\tau_2^{-1}\tau_1(i)=(i)$ .
I think it is not correct. For instance, take $n=6$ and let $\tau_1=(1\ 2)(3\ 4\ 5\ 6),\tau_2=(1\ 2\ 3)(4\ 5\ 6)$ . Clearly $\tau_1(1)=\tau_2(1)=2$ but $\tau_1\ne\tau_2$ . Am I misunderstanding it?
The whole text is below:
