Alternating groups, $A_3$ and $A_4$

Question

I have to:

1. Prove that $A_3$ is a simple group.

2. Prove that $A_4$ is not a simple by proving that the elements of order $2$ along with the neutral form a normal subgroup.

I know that the elements of $A_3$ are $\{1,(123),(132)\}, $ but how is $A_3$ a simple group then?

Answer 1

$A_3$ is isomorphic to $\Bbb{Z}_3$ since the order is $3$. Thus there are no other proper non-trivial normal subgroups. Hence $A_3$ is a simple group. In general, $\Bbb{Z}_p$ is simple for any prime $p$.