Show that the alternating group $A_{n}$ is nonabelian for $n\geq 4$.

Question

How would I show this? Do I use induction? I am not sure how to even begin with induction, and so I have no work to show.

Answer 1

We don't even need induction. If $k \leq n$, then $A_k$ is a subgroup of $A_n$, so it suffices to treat the case $k=4$.

We have $(1,2,3)\cdot(2,3,4)=(1,2)(3,4)$, but $(2,3,4)\cdot (1,2,3)= (1,3)(2,4)$, so $A_4$ is not abelian.