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I'm trying to prove that $G := GL_3(\mathbb{F}_2)$, the group of $3 \times 3$ matrices with entries in $\mathbb{F}_2$ is a simple group. The steps outlined for me look like:

1) Construct a list of representatives for the conjugacy classes of $G$.

2) Compute the size of each of these conjugacy classes.

3) Show that $G$ is simple.

I was able to solve step (1) using the fact that every matrix in $G$ is conjugate to a unique block matrix, where each of the blocks are companion matrices of a list of invariant factors. That is, for each matrix $A \in G$ there exists unique (up to associates) $\delta_1 \mid \cdots \mid \delta_n$, $\delta_i \in \mathbb{F}_2[x]$ such that $A \sim $diag(Com($\delta_1$), $\ldots$, Com($\delta_n$)). Using this fact I was able to construct the following list of representatives for conjugacy classes of matrices: $$ \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} , \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix} , \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

I get a little stuck on part (2). I know the size of the conjugacy class of each of the above matrices is equal to the index of the centralizer. I could compute the centralizer of each of the above matrices directly and that would give me the answer but I'm a little hesitant to just multiply matrices for 10-15 minutes. This problem was on a practice qualifying exam so I suspect that there is a faster/more clever way to compute the sizes of these conjugacy classes. This is really what I want. One idea I have:

Two matrices are conjugate if and only if they have the same list of invariant factors. For many of the matrices the list of invariant factors is a single degree 3 polynomial. In this case I know both the minimal polynomial and characteristic polynomial of any matrix conjugate to my representative. These observations do not seem to make the computation much faster though.

I suspect that once I can do (2), (3) will follow relatively quickly.

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Calling your six matrices $a, b, c, d, e, f$ you can find their centralizers by looking at the action of $G$ on the seven non-zero vectors in $\mathbb{F}_2^3$:

$a$ maps 100 to 010, 010 to 001 and 001 to 100 giving an orbit of length $3$. Another orbit of length 3 is $\{011, 101, 110\}$ and there is the fixed point $111$. As permutation $a$ is the product of two $3$-cycles. In the symmetric group $S_7$ its centralizer looks like $Z_3^2\rtimes Z_2$, as its elements have to respect the orbits of $\langle a\rangle$ by either fixing them (mapping its elements back into the orbit, no elementwise fixing) or exchanging orbits of the same size. Take a centralizing element $x$ that exchanges the orbits, i.e., $xa=ax$ and let's say without loss of generality that $x\cdot100=110$ (otherwise replace $x$ by $ax$ or $a^2x$). Then $x\cdot010 = xa\cdot100=ax\cdot100=a\cdot110=011$, and so by linearity of $x$ we get $x\cdot110=x\cdot100 + x\cdot010=110 + 011=101$ contradicting that $x$ exchanges both orbits of length $3$. A linear permutation fixing the orbit $\{100,010,001\}$ has to be a power of $a$, showing that $\langle a\rangle$ is self-centralizing in $G$.

Alternatively you can prove that an element $x$ of the centralizer of $a$ has to map each orbit of order $3$ to itself by observing that $\{011,101,110\}$ plus the zero vector is a subspace of dimension $2$ whose image under a linear $x$ has to intersect itself non-trivially. The restriction of $x$ to the other orbit (i.e., viewing its image in $S_3$) has to be a power of $a$, but as this orbit is a basis of the vector space, a linear $x$ has to be a power of $a$.

$b$ and $d$ have orbits of length $7$, and have as $7$-cycles already in $S_7$ the centralizers $\langle b\rangle$ rsp. $\langle d\rangle$.

$c$ has orbits $\{100, 010, 001, 111\}$, $\{011, 110\}$ and fixed point $101$. An element $x\in S_7$ centralizing $c$ has to map the orbit of length $4$ to itself, and restricted to this orbit be a power of $c$. As the orbit contains a basis of the vector space, a linear $x$ is therefor a power of $c$.

As you now know the sizes of the conjugacy classes of $a,b,c,d,f$ you skip this kind of argument for $e$, as its conjugacy classes consists of the remaining elements of $G$.

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