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Exercise :

Give all the non isomorphic abelian groups of order $225$ and collapse them in to the minimum possible.

Attempt :

We have that the prime factorization of $225$ is :

$$225 = 3^2\cdot5^2$$

This means that all the non isomorphic abelian groups of order $225$ are :

$$\mathbb Z_{3^2} \times \mathbb Z_{5^2} $$

$$\mathbb Z_{3} \times \mathbb Z_{3} \times \mathbb Z_{5^2}$$

$$\mathbb Z_{3^2} \times \mathbb Z_{5} \times \mathbb Z_{5}$$

$$\mathbb Z_{3} \times \mathbb Z_{3} \times \mathbb Z_{5} \times \mathbb Z_{5}$$

How would one proceed by collapsing (shortening) one of the above expressions to the shortest (minimum) expression ?

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  • $\begingroup$ You have to write them as products of quotients by the invariant factors. $\endgroup$ – Bernard Jan 21 '18 at 12:01
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You have used Fundamental theorem of abelian groups .So,after that, the shortening using that $gcd(a,b)=1$ is :

1)$\mathbb Z_{3^2} \times \mathbb Z_{5^2}\cong\mathbb Z_{225} $

2)$\mathbb Z_{3} \times \mathbb Z_{3} \times \mathbb Z_{25} \cong\mathbb Z_{3}\times \mathbb Z_{75} $

3)$\mathbb Z_{9} \times \mathbb Z_{5} \times \mathbb Z_{5} \cong\mathbb Z_{45}\times \mathbb Z_{5}$

4)$\mathbb Z_{3} \times \mathbb Z_{3} \times \mathbb Z_{5}\times \mathbb Z_{5} \cong\mathbb Z_{15}\times \mathbb Z_{15}$

Ps: $\mathbb Z_{a} \times \mathbb Z_{b} $

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