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.


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