# Find invariant factors from elementary divisors of an abelian group of order 720

For any abelian group of order $$720=2^43^25^1$$, by writing out all the partitions of each of the powers $$4,2,1$$. I found that the abelian group must be isomorphic to only one of the following: $$\mathbb Z_{2^4} \times \mathbb Z_{3^2} \times \mathbb Z_{5^1} \\ \mathbb Z_{2^4} \times (\mathbb Z_{3^1} \times \mathbb Z_{3^1}) \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^3} \times \mathbb Z_{2^1}) \times \mathbb Z_{3^2} \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^3} \times \mathbb Z_{2^1}) \times (\mathbb Z_{3^1} \times \mathbb Z_{3^1}) \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^2} \times \mathbb Z_{2^2}) \times \mathbb Z_{3^2} \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^2} \times \mathbb Z_{2^2}) \times (\mathbb Z_{3^1} \times \mathbb Z_{3^1}) \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^2} \times \mathbb Z_{2^1} \times \mathbb Z_{2^1}) \times \mathbb Z_{3^2} \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^2} \times \mathbb Z_{2^1} \times \mathbb Z_{2^1}) \times (\mathbb Z_{3^1} \times \mathbb Z_{3^1}) \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^1} \times \mathbb Z_{2^1} \times \mathbb Z_{2^1} \times \mathbb Z_{2^1}) \times \mathbb Z_{3^2} \times \mathbb Z_{5^1} \\ (\mathbb Z_{2^1} \times \mathbb Z_{2^1} \times \mathbb Z_{2^1} \times \mathbb Z_{2^1}) \times (\mathbb Z_{3^1} \times \mathbb Z_{3^1}) \times \mathbb Z_{5^1}$$ I am trying to find the invariant factors.

My vague understanding is that I should find the elementary divisors first, so that I can group them together by powers of the same prime numbers and then multiply them accordingly. I think my elementary divisors are $$2^1,2^2,2^3,2^4,3^1,3^2,5^1$$; when rearranging them in a square grid, that is, $$2^1,2^2,2^3 \\ 3^0,3^1,3^2 \\ 5^0,5^0,5^1$$ can I multiply downward so that I can get the invariant factors of $$2^13^05^0,2^23^15^0,2^33^25^1$$? I think these are invariant factors because my first one divides my second, which in turn divides my third.