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 $A_4$. 

Hints


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*To prove $d(G) \le k$, find $k$ elements of $G$ that generate $G$.

*To prove $d(G) \ge k$, prove that any set of $k-1$ elements will not generate $G$. In particular, when $k=2$, what does this statement translate to?

