Confusion in understanding of Steinitz Exchange Lemma by an example

Question

I am trying to understand Steinitz Exchange Lemma from Kurosh's Group Theory:

Suppose in a group $G$, two finite systems of elements are given: $$ u', u'', \ldots, u^{(k)} \hskip5mm (I) \hskip1cm \mbox{ and } \hskip5mm v', v'', \ldots, v^{(l)} \hskip5mm (II) $$ the first of which is linearly independent system, each of whose elements is linearly dependent on the second system. Then $k\le l$ and from $(II)$, $k$ elements can be omitted such that the remaining elements with the elements of $(I)$ form a system equivalent to $(II)$.

My question is simple, but I am not getting my fault in understanding.

Suppose we have group $G=\mathbb{Z}\oplus \mathbb{Z}$, with following systems of elements. $$ (2,0), (0,2) \hskip5mm (I) \hskip1cm \mbox{ and } \hskip1cm (1,0), (0,1)\hskip5mm (II). $$ It is clear that

• $(I)$ is linearly independent (their $\mathbb{Z}$-linear combination is zero only when coefficients are zero).

• Members of $(I)$ are linearly dependent on $(II)$ (obvious).

• $k\le l$ (they are equal to two both).

As per assertion, we can omit two elements from $(II)$ and replacing them two elements from $(I)$, we get a system equivalent to $(II)$; it means that $(I)$ and $(II)$ should be equivalent. But this is not the case - members of $(II)$ are not $\mathbb{Z}$-combinations of members of $(I)$.

Can anybody point out my misunderstanding?

Answer 1

According to Kurosh, systems (I) and (II) are equivalent.

Note the definition given by Kourosh: An element $u$ of $G$ linearly dependent on the system of elements $u_1,u_2,\ldots,u_l$ of $G$ if

$$ \alpha u=\beta_1 u_1+\beta_2 u_2+\ldots+\beta_lu_l,\ \alpha,\beta_i\in\mathbb{Z}. $$

And perhaps $\alpha\neq1$.