Upper and lower bounds for the determinant of the discrete subgroup of $\mathbb R^n$. Let $\Delta \subset \mathbb R^n$ be a discrete subgroup generated by $\mathbb R$-linear independent basis $\{x_1,\dots, x_r\}$ (namely $\Delta$ is an Abelian group of rank $r$).
The determinant of $\Delta$, denoted $d(\Delta)$, is defined to be the absolute value of the determinant of $\tau$ where $\tau$ is a linear transformation on the $\mathbb R$-linear space $\text{Span}\{x_1,\dots, x_r\}$ such that $\{\tau^{-1} x_1,\dots, \tau^{-1} x_r\}$ becomes an orthonormal basis of the $\mathbb R$-linear space $\text{Span}\{x_1,\dots, x_r\}$. 
Now let $g$ be a nonsingular matrix, then I wonder how to prove the following proposition:
There exist positive constants $a$ and $b$ such that $ad(\Delta)\le d(g\Delta) \le bd(\Delta)$ for any discrete subgroup $\Delta \subset \mathbb R^n$.
There is a fact which may be helpful:
The determinant of the Gram matrix $A:=(\langle x_i, x_j \rangle)_{ij}$ turns out to be the square of the determinant of $\Delta$. But even with this, I found it hard to prove the existence of such $a$ and $b$.
 A: Let $V$ be a finite-dimensional real vector space equipped with the inner product $\langle \cdot , \cdot \rangle$. Consider the $r$-fold exterior power of $V$: $W=\Lambda^r V$. The standard inner product $\langle \cdot , \cdot \rangle$ on $W$ induced from the one on $V$ is uniquely determined by its values on pairs of decomposable alternating $r$-tensors:
$$
\langle x_1\wedge ... \wedge x_r , y_1\wedge ... \wedge y_r \rangle 
=\det(M),
$$ 
where $M$ is the matrix with the $ij$-entry $\langle x_i, y_j\rangle$. See for instance this  Wikipedia page. 
In particular, for $w= x_1\wedge ... \wedge x_r$ the norm (with respect to the standard inner product on $W$) equals the square root of the determinant of the Gramm matrix of the tuple $\bar x=(x_1,...,x_r)$, i.e. equals  $d(\bar x):=d(\Delta)$ in your notation, as was observed in a comment by Alex Ravsky. Here $\Delta$ is the discrete subgroup generated by the vectors in $\bar x$. Personally, I prefer to think of $d(\Delta)$ as the covolume of $\Delta$, i.e. the volume of the quotient of the unique $\Delta$-invariant linear subspace on which $\Delta$ acts cocompactly, i.e. the linear span of $\{x_1,...,x_r\}$. 
Now, let $A\in GL(V)$ be an invertible endomorphism of $V$. It acts naturally on $W$, this action  is denoted by $\Lambda^r(A)$ and satisfies:
$$
\Lambda^r(A)(x_1\wedge ... \wedge x_r)= Ax_1\wedge ... \wedge Ax_r, 
$$
and $\Lambda^r(A^{-1})= (\Lambda^r(A))^{-1}$. 
As any finite-dimensional linear operator, $\Lambda^r(A)$ has finite operator norm $||\Lambda^r(A)||$ with respect to the the standard norm on $W= \Lambda^r V$.  Then for every $w\in W$ we have 
$$
||\Lambda^r(A^{-1})||^{-1} \cdot ||w|| \le ||\Lambda^r(A) w|| \le ||\Lambda^r(A)||\cdot ||w||.
$$
In the special case of decomposable alternating tensors $w= x_1\wedge ... \wedge x_r$, we get the inequality that you are after:
$$
||\Lambda^r(A^{-1})||^{-1} d(x_1,...,x_r) \le d(Ax_1,...,Ax_r) \le ||\Lambda^r(A)|| d(x_1,...,x_r),
$$
equivalently, for the discrete subgroup $\Delta$ generated by $x_1,...,x_r$
$$
||\Lambda^r(A^{-1})||^{-1} d(\Delta) \le d(A\Delta) \le ||\Lambda^r(A)|| d(\Delta). 
$$
Lastly, one can compute the norm $||\Lambda^r(A)||$ explicitly: If
$$
\sigma_1\ge \sigma_2\ge ... \ge \sigma_n>0, 
$$
are the singular values of $A\in GL_n({\mathbb R})$ (counted with multiplicity), then $||\Lambda^r(A)||= \sigma_1\cdots \sigma_r$. To prove this formula, observe that the SVD gives you $A=UDV$, where $U, V\in O(n)$ and $D=Diag(\sigma_1,...,\sigma_n)$. Since $U, V$ acts isometrically on $W$, 
$$
||\Lambda^r(A)||= ||\Lambda^r(D)||.
$$
The action of $D$ on $W$ is diagonalizable with the orthogonal eigenbasis given by the tensors of the form
$$
e_{i_1}\wedge e_{i_2}\wedge ... \wedge e_{i_r},
$$
$i_1<i_2<...<i_r$ and their respective eigenvalues equal
$$
\sigma_{i_1} ... \sigma_{i_r}.
$$ 
From this, you see that the highest eigenvalue is exactly $\sigma_1\cdots \sigma_r$. 
