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I have met this strange looking problem on which I have no idea, from my course on Abstract Algebra dealing with modules:

Let $ v_1,...,v_k \in \mathbb{Z}^n $ row vectors of length n over $ \mathbb{Z} $

and we denote the subgroup of $ \mathbb{Z}^n $ generated by these vectors by $ (v_1,...,v_k)= \mathbb{Z}v_1 + ... + \mathbb{Z}v_k $. Now we define the following matrix: $ A \in M_{k \times n}(\mathbb{Z}) $ the matrix whose ith row is the vector $ v_i $, and we are now given a matrix $ g \in GL_k(\mathbb{Z}) $. We denote by $ w_1,...,w_k $ the rows of the matrix $ gA $, we are asked to show that $ (v_1,...,v_k) = (w_1,...,w_k) $ i.e. the subgroups spanned by the rows of A and gA over $ \mathbb{Z} $ are equal.

I know Abelian groups can be thought of as $ \mathbb{Z} $ modules but I really do not know how to proceed, I certainly need help to solve this and thank all helpers

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For $1 \le i \le k$, we have $w_i = \sum_{j=1}^k g_{ij}v_j$. So each vector $w_i$ is an integral linear combination of the vectors $v_j$, and hence $(w_1,\ldots,w_k) \le (v_1,\ldots,v_k)$.

If $B \in M_{k \times n}({\mathbb Z})$ denotes the matrix whose rows are the $w_i$, then $B = gA$, so $A = g^{-1}B$ and we get the reverse inclusion.

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