Distributive subgroups lattice

Question

Let $G = \langle a \rangle \times \langle b \rangle $ ($a,b \in G$), where $ |\langle a \rangle| = n, |\langle b \rangle| = m$ and $gcd(n,m) = d > 1$. I need to show that subgroup lattice of $G$ is not distributive. I want to show that $ \langle ab \rangle \cap (\langle a \rangle \cup \langle b \rangle) = \langle ab \rangle \neq (\langle ab \rangle \cap \langle a \rangle) \cup( \langle ab \rangle \cap \langle b \rangle)$, becouse it works, if n or m is infinity, but i don't undestand how find $\langle ab \rangle \cap \langle a \rangle$ and $\langle ab \rangle \cap \langle b \rangle$

Answer 1

Hint: $\langle ab \rangle \cap \langle a \rangle =\{a ^ i: m | i\}$ because $(ab) ^ i = a ^ i\implies b ^ i = 1\implies m | i $.