“Redundant” relations in Finitely Generated Abelian Groups

Let $$G=\langle x,y,z\rangle/\langle2y-x, 4z-2x, 2z-2y\rangle$$ denote the free abelian group generated by $$\{x,y,z\}$$ quotiented out by the subgroup generated by $$\{2y-x,4z-2x,2z-2y\}$$.

Let $$H=\langle x,y,z\rangle/\langle 2y-x,2z-2x\rangle$$. We note that $$G=H\cong Z\oplus Z/2Z$$, mainly because the relation "$$4z=2x$$" is "redundant" in the sense that it follows from $$2y=x$$ and $$2z=2y$$.

However, $$J=\langle x,y,z\rangle/\langle 4z-2x,2z-2y\rangle\cong Z\oplus Z/2Z\oplus Z/2Z$$ is not isomorphic to $$G$$. This is because the missing relation "$$2y=x$$" cannot be deduced from the other two relations.

I am interested to know the underlying reason and theory behind this phenomenon (in general). Given an abelian group with finite many generators and relations, can we tell which relations are "redundant". I am vaguely aware it is related to Smith Normal Form, can someone point me in the right direction of the related theorems and results? I

Is it possible to find out "redundant" relations without performing the entire Smith Normal algorithm? Any partial/sufficient conditions for finding "redundant" relations will be helpful too.

Thanks a lot.

• The relators span a subgroup of the free abelian group spanned by the generators. So you are asking whether one of the relators is in the subgroup generated by the others. The Hermite Normal Form provides a canonical free basis for a subgroup of a finitely generated free abelian group, so that might be more useful than the Smith Normal Form. From the canonical basis, you can easily decide whether a given element is in the subgroup. – Derek Holt Sep 21 '18 at 8:11
• @DerekHolt Great. Thanks. Is there any good book or reference for Hermite Normal Form? – yoyostein Sep 21 '18 at 12:31