Lagrange theorem question I'm trying to teach myself group theory and this question is the final one in an exercise on Lagrange theorem and it has me currently stumped.  
Finite group ${G}$ contains distinct elements ${a}$ and ${b}$ and identity ${e}$, such that: ${a*b=a^3*b^4}$ and ${a^4*b^3=e}$. Show that ${a^2=e}$ and that the order of ${G}$ is a multiple of 6.

1st question : $a^4*b^3=e \Rightarrow b^3 = a^{-4}$ and
$$a*b = a*a^2*b^3*b \Rightarrow e=a^2*b^3 = a^2 * a^{-4} = a^{-2}$$
so $a^2 = (a^{-2})^{-1} = e$.
