Isomorphism for invertible diagonal matrix

Question

We know that the set of invertible diagonal $n\times n$ matrix over $\mathrm{GF}(q)$ ($\mathrm{DL}(n,q)$ is a subgroup of $\mathrm{GL}(n,q)$). Is there any isomorphism for $\mathrm{DL}(n,q)$? I think $\mathrm{DL}(n,q)$ is isomorphic to product of $\mathrm{GF}(q)^\times$ $n$ times to itself. But I can't prove it. Please help me.

Answer 1

In fact, if $K$ is any field, then the subgroup of diagonal matrices of $\mathrm{GL}(n,K)$ is isomorphic to $(K^*)^n$. The isomorphism maps $(\lambda_1,\dotsc,\lambda_n) \in (K^*)^n$ to the diagonal matrix $\mathrm{diag}(\lambda_1,\dotsc,\lambda_n)$.

Answer 2

You are correct. The diagonal matrix $$ \operatorname{diag}(d_1, \ldots, d_n) = \begin{bmatrix} d_1 & & \\ & \ddots & \\ & & d_n \end{bmatrix} \in M_n(GF(q)) $$ is invertible if and only if each $d_i \in GF(q)^\times$ since $$ \operatorname{diag}(d_1, \ldots, d_n) \operatorname{diag}(d'_1, \ldots, d'_n) = \operatorname{diag}(d_1 d'_1, \ldots, d_n d'_n). $$ Now, the map $$ \begin{align} DL(n, q) &\to GF(q)^\times \times \cdots \times GF(q)^\times \\ \operatorname{diag}(d_1, \ldots, d_n) &\mapsto (d_1, \ldots, d_n) \end{align} $$ is the required isomorphism. Note that this argument works for any field.