general linear group $GL_{2}(\mathbb{Z}_3).$ finding the order of the general linear group $GL_{2}(\mathbb{Z}_3).$ ?
Is finding the order by listing its elements only ? or is there a smarter method?
 A: Just count the possible columns of a matrix in $GL_2(\Bbb{Z}_3)$. The first column has to be nonzero, which gives $9 -1 = 8$ possibilities. The second column should not be a multiple of the first column, which gives $9 - 3 = 6$ possibilities. Thus you find $8 \times 6 = 48$ elements.
A: Here are two approaches, both thinking of $GL_2(\Bbb Z_3)$ as the group of $2\times 2$ invertible matrices.

*

*Count the number of elements in $M_{2\times2}(\Bbb Z_3)$ which are rank 0 and rank 1, and conclude how many have rank 2

*Count the number of possible first columns, then count the number of possible second columns

A: Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\text{GL}_2(\Bbb Z_3)$. Then $\det(A)\neq 0$. So you need to solve $ad-bc\neq 0$. A fast way is to see that there are $3^4$ different matrices, so you just have to substract the solutions for $ad-bc=0$.
A: Let me try and sketch the general argument which allows the evaluation of the order of general linear groups of finite dimensional spaces. Let $K$ be an arbitrary field (not necessarily commutative) and $V$ be a left $K$-vector space of finite dimension $\mathrm{dim}_KV=n \in \mathbb{N}$. For each $k \leqslant n$ define $L_k$ to be the subset of $V^k$ consisting of all linearly independent sequences of vectors.

*

*The linear group $\mathrm{GL}_K(V)$ has a natural (componentwise) action on each set $L_k$, all of these actions being transitive (this claim relies on the fact that any linearly independent subset/family can be extended to a basis). Furthermore, the action on $L_n$ is regular (i.e. not only is it transitive but all the elements are free, of trivial stabilisers; this is due to the fact that any morphisms that agree on a generating system are equal). Thus you can draw the conclusion that $|\mathrm{GL}_K(V)|=|L_n|$ (there are as many automorphisms as there are ordered bases).

*For each $k<n$ there exists a natural restriction map $\rho_k \colon L_{k+1} \to L_k$, which to any sequence $x$ of linearly independent vectors indexed by the natural interval $[1, k+1]$ assigns its restriction to the subinterval $[1, k]$ (disregarding the "last" component). The fibre of an element $y \in L_k$ is described as:
$$\rho_k^{-1}[\{y\}]=\left\{x \in L_{k+1}|\ (\forall i)(i \leqslant k \Rightarrow x_i=y_i) \wedge x_{k+1} \in V \setminus \left\langle\{y_i\}_{1 \leqslant i \leqslant k}\right\rangle_K\right\},$$
where by $\langle X \rangle_K$ I am referring to the $K$-subspace of $V$ generated by the subset $X \subseteq V$. Since $y$ is a linearly independent family, the subspace its components generate is $k$-dimensional and therefore its complementary has the cardinality of the complementary of a $k$-dimensional $K$-vector space within an $n$-dimensional vector space. It is easy to see that the cardinality of such a complementary is given by $q^n-q^k$ when $K$ is a finite field of cardinality $|K|=q \in \mathbb{N}$ respectively by $|K|$ itself when $K$ is infinite.

*We return to the decomposition of $L_{k+1}$ as a disjoint union of subsets (the fibres of $\rho_k$) all of which have an equal cardinality -- the one discussed above -- and which are in number of $|L_k|$. This establishes the recursive relation $|L_{k+1}|=\left(q^n-q^k\right)|L_k|$ in the case $K$ finite respectively $|L_{k+1}|=|K||L_k|$ in the case $K$ infinite. Since $L_0$ is a singleton (consisting of the empty family), we conclude that:
$$|\mathrm{GL}_K(V)|=|L_n|=
\begin{cases}
\displaystyle\prod_{k<n}\left(q^n-q^k\right), &|K|=q \in \mathbb{N}\\
|K|, &|K|\notin \mathbb{N} \wedge n \neq 0\\
1, &|K|\notin \mathbb{N} \wedge n=0.
\end{cases}$$
In your particular case, since $\mathrm{GL}_2(\mathbb{Z}_3) \approx \mathrm{GL}_{\mathbb{Z}_3}\left(\mathbb{Z}_3^2\right)$ as groups we have $q=3$ and $n=2$ in the above notations so therefore:
$$|\mathrm{GL}_z(\mathbb{Z}_3)|=\left(3^2-1\right)\left(3^2-3\right)=48.$$
A: We know $M_2(\mathbb{Z}_3)\cong \mathbb{Z}_3^4$ is of order 81. Since $\mathbb{Z}_3$ is a field, One idea would be to write down the characteristic polynomial of a general element and see when it has zero as a root. Elements whose characteristic polynomial has zero as a root are precisely the non-invertible matrices. The rest are precisely $GL_2(\mathbb{Z}_3)$.
Writing down the elements might be faster, though.
