Minimum Number of Generators A subset $S$ is said to generate a group $G$ if every element of $G$ can be written as a finite product of elements of $S$ and their inverses. Let $d(G)$ be the minimum number of generators of $G$. 
Prove that the quaternion group of order 8, $Q_8$ satisfies $d(Q_8)=2$ and that the alternating group $A_4$ satisfies $d(A_4)=2$. 
My approach: Well $Q_8$ is generated by $(i,k), (j,k)$ or $(i,j)$. So this is easy to see. I am not quite sure about $A_4$. 

$Q_8$ is not cyclic, so we need at least two elements, consider $i$ and $j$, see that generates.
$A_4$ is also not cyclic, prove it is generated by $(1,2,3)$ and $(1,2,4)$.
