# How can I determine if the group $G=(\Bbb Z_{16}\times\Bbb Z_{18},+)$ is isomorphic to $(\Bbb Z_4\times\Bbb Z_{72},+)$?

How can I determine if the group $$G=(\Bbb Z_{16}\times\Bbb Z_{18}, +)$$ is isomorphic to $$(\Bbb Z_4\times\Bbb Z_{72}, +)$$?

I came up with that $$(1,1)$$ in $$G$$ has order $$144$$ because $$\operatorname{lcm}(16,18)=144$$.

Highest possible order for an element in $$(\Bbb Z_4\times\Bbb Z_{72}, +)$$ is $$72$$ because $$\operatorname{lcm}(4,72)=72$$.

Therefore they cant be isomorphic because there is no element with order $$144$$ in $$(\Bbb Z_4\times\Bbb Z_{72}, +)$$.

Is there any other way I can prove it?

You can invoke the fundamental theorem of finite abelian groups. Decompose $$\mathbb{Z}_{16} \times \mathbb{Z}_{18}\cong \mathbb{Z}_{2^4} \times \mathbb{Z}_2 \times \mathbb{Z}_{3^2}$$ $$\mathbb{Z}_{4} \times \mathbb{Z}_{72}\cong \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{2^3}$$ and such a decomposition is unique (up to permutation of the factors). Since the decompositions do not agree, the groups are not isomorphic.
Here, I used the fact that $$\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm}$$ when $$m$$ and $$n$$ are coprime.