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.