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If I want to simplify the group presentation (free abelian group with relations)

$$\langle a,b,c\mid 2a=b=2c\rangle,$$

I can simplify it as $$\langle a,c\mid 2a=2c\rangle\cong\langle a,a-c\mid 2(a-c)\rangle\cong\mathbb{Z}\oplus\mathbb{Z}_2.$$

How about for more complicated presentations such as $$\langle a,b,c\mid 2a=3b=5c\rangle?$$

How do I simplify it, if possible?

In general, I am interested in simplifying $$\langle a,b,c\mid n_1a=n_2b=n_3c\rangle,$$ where $n_1,n_2,n_3\in\mathbb{Z}$.

Thanks.

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    $\begingroup$ The general method for this problem is the Smith Normal Form diagonlization algorithm for matrices over the integers. $\endgroup$ – Derek Holt Feb 5 '18 at 9:46

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