Number of orbits and representatives of an action of $Gl_2(\mathbb{Z}_2)$ in $M_2(\mathbb{Z}_2)$ by conjugation

Question

So I'm asked to find the number of orbits and representatives of that action, my idea was to find all the possible rationals forms induced by a polynomial of order 2 in $\mathbb{Z}_2[x]$, thinking of it as modules over $\mathbb{Z}_2[x]$, but i only seem to find 6, with the modules by $\mathbb{Z}_2[x]/x²,\mathbb{Z}_2[x]/(x-1) \bigoplus \mathbb{Z}_2[x]/x,\mathbb{Z}_2[x]/(x-1)²,\mathbb{Z}_2[x]/(x²+x+1),\mathbb{Z}_2[x]/(x-1)\bigoplus\mathbb{Z}_2[x]/x ,\mathbb{Z}_2[x]/x \bigoplus \mathbb{Z}_2[x]/x$, but they are supposed to be 7 according to the solutions, can u help me out? Thanks.

Answer 1

Describing the orbits together with representatives. Recall that $|GL_2(\Bbb{F}_2)|=6$. As it acts faithfully on the set of three non-zero vectors of the space $\Bbb{F}_2^2$ we see that it must be isomorphic to $S_3$. This helps in what follows.

• The group $G=GL_2(\Bbb{F}_2)\simeq S_3$ has three conjugacy classes. Determined by the cycle type of the permutation. Representatives: $$I_2,\qquad\left(\begin{array}{cc}0&1\\ 1&0\end{array}\right),\qquad\left(\begin{array}{cc}0&1\\ 1&1\end{array}\right).$$
• The unique rank zero matrix, $0_{2\times2}$, forms a singleton orbit. This leaves $2^{2\times2}-6-1=9$ matrices of rank one.
• The nilpotent rank one matrices form a single orbit. When viewed as linear transformations these have a single non-zero vector in their image. It is obvious that conjugating the transformation by a prescribed permutation of the three non-zero vectors lets us choose which non-zero vector is in the image. Thus there are 3 matrices in this orbit. A representative of this orbit is $$\left(\begin{array}{cc}0&1\\0&0\end{array}\right).$$
• The remaining rank one matrices are all idempotent. As a linear transformation such a matrix is uniquely determined if we know which non-zero vector is in the kernel and which other non-zero vector is in the image. The action of $G$ is doubly transitive, so all six combinations are in the same $G$-orbit. A representative of this orbit is $$\left(\begin{array}{cc}1&0\\0&0\end{array}\right).$$

The OP used the classification method of identifying the $R=\Bbb{F}_2[x]$-module structure given to the space $\Bbb{F}_2^2$ by letting the indeterminate $x$ act via the chosen matrix. The six orbits above correspond to modules $(R/\langle(x+1)\rangle)^2$, $R/\langle x^2+1\rangle$, $R/\langle x^2+x+1\rangle$,$(R/\langle x\rangle)^2$, $R/\langle x^2\rangle$ and $R/\langle x\rangle\oplus R/\langle x+1\rangle$ respectively (in the above order).

Anyway, the sixteen $2\times2$ matrices are partitioned into six orbits of the conjugation action of $G$.