Taking a basis from an old basis in Free Abelian Groups

Question

Let $G$ be a free abelian group with free generators $x_1,x_2,x_3$. We want to show that $\forall g:=(g_{ij})\in \mathrm{GL_3}(\mathbb{Z})$, the elements $u_j=\sum_{i=1}^{3} g_{ij}x_i,\ j=1,2,3$ are also free generators of $G$.

My first thought is to show that the elements $u_j$ are $\mathbb{Z}-$linearly independent and generating $G$. But this is not helpful, and I have definitely stuck.

Any ideas please?

Thank you.

Answer 1

Let $h=g^{-1}$ be the inverse of $g$ in the linear group. Then we have a corresponding relation describing each $x_i$ in terms of $u_1,u_2,u_3$. (In the post, the sum is over $i$.)

From this, $u_1,u_2,u_3$ are also generators. One can formally write: $$ \begin{bmatrix} u_1\\u_2\\u_3 \end{bmatrix} = \underbrace{ \begin{bmatrix} g_{11}&g_{12}&g_{13}\\ g_{21}&g_{22}&g_{23}\\ g_{31}&g_{32}&g_{33} \end{bmatrix} }_{:=g} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}\ ,\qquad \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = \underbrace{ \begin{bmatrix} h_{11}&h_{12}&h_{13}\\ h_{21}&h_{22}&h_{23}\\ h_{31}&h_{32}&h_{33} \end{bmatrix} }_{:=h} \begin{bmatrix} u_1\\u_2\\u_3 \end{bmatrix}\ , $$ If $v$ is an elements, then it is generated by the $x$ basis, formally written, we find "$a$-scalars" such that $$ v = \underbrace{ \begin{bmatrix} a_1&a_2&a_3 \end{bmatrix}}_{a} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}\ . $$ Then $$v = \underbrace{ \begin{bmatrix} a_1&a_2&a_3 \end{bmatrix} \begin{bmatrix} h_{11}&h_{12}&h_{13}\\ h_{21}&h_{22}&h_{23}\\ h_{31}&h_{32}&h_{33} \end{bmatrix} }_{:=b} \begin{bmatrix} u_1\\u_2\\u_3 \end{bmatrix}\ , $$ so $v$ is also generated by the $u$-system, use the components of $b$ above.

Let us now assume that we have a linear combination of the $u$-system, which is zero: $$ 0 = \underbrace{ \begin{bmatrix} a_1&a_2&a_3 \end{bmatrix}}_{a} \begin{bmatrix} u_1\\u_2\\u_3 \end{bmatrix}\ . $$ Then $$0 = \underbrace{ \begin{bmatrix} a_1&a_2&a_3 \end{bmatrix} \begin{bmatrix} g_{11}&g_{12}&g_{13}\\ g_{21}&g_{22}&g_{23}\\ g_{31}&g_{32}&g_{33} \end{bmatrix} }_{:=b} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}\ , $$ so $b=0$, since $x$ is a basis, so $a =agh=bh=0h=0$. This shows that the system $u$ is independent.