If $\langle a \rangle \cap \langle b \rangle=\{e\}$ then the group contains one element with least common multiple of m,n.
I have constructed the element $ab$ will have order $ab=LCM(m,n)$.
$$ LCM(m,n)= fm= gn:1\le f\le n,1\le f\le m $$
$$ab^{LCM(m,n)}=(a^{m})^f(b^n)^g= e$$
Is this proof correct I used commutative nature to get that exponent distributes but I never seem to use the fact that intersection is just identity. Where have I gone wrong?
Best Answer
To finish it off:
$(ab)^r=e\implies a^r=b^{-r}\implies a^r=e\implies m\mid r$. Similarly, $n\mid r$. $\therefore r\ge\operatorname{lcm}(m,n)$.