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.

Proof that $a^2=e$:
$a^4*b^3=e$. Multiply both sides by $b$ from right to get to $a^4*b^4=b$. This combined with the assumption that $a*b=a^3*b^4$, implies $a*a*b=b$ (more details: $a^4*b^4=a*a^3*b^4=a*a*b$). Multiply both sides by $b^{-1}$ to get to $a*a=e$ or $a^2=e$.
Proof that $|G|$ is a multiple of $6$: 
First we prove that $b^3=e$.
We know that $a*b=a^3*b^4$ and $a^2=e$, hence $a*b=a^2*a*b^4=a*b^4$. So by multiplying both sides by $a^{-1}$ from left and then by $b^{-1}$, we get to $b^3=e$.
By now, we know that $G$ has an element of order $2$ and an element of order $3$. It means that $|G|$ must be a multiple of $2$ and $3$. Hence $|G|$ is a multiple of $6$ nad we're done!
