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I'm asked to find out all of the conjugacy classes, their order and their size for $GL(4,2)$.

Finding representatives is possible by looking for all the rational canonical forms over the field and the order, just by taking powers of the representatives.

Now, about calculating the size of each class, I know it can be done by trying to calculate the size of the center for each class, $C_G(x_i)$, where $x_i$ are the representatives, and then $\frac{|GL(4,2)|}{|C_G(x_i)|}$ is the size of the class, but it appears to be very difficult to do it straightforward.

Can someone suggest me a better way to do it?

Thanks

Note: $GL(4,2)$ is all invertible matrices of size $4\times4$ over $F_2$. http://en.wikipedia.org/wiki/General_linear_group#Over_finite_fields

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Well, ${\rm GL}(4,2) \cong A_8$. I am not sure if that is very helpful without an explicit isomorphism, but it might at least help you check your answer! –  Derek Holt Feb 14 '13 at 18:22
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Sometimes work needs to be done on questions like this. –  Geoff Robinson Feb 14 '13 at 18:24
    
Just to clarify, is $GL(4,2)$ the general linear group of a 4-dimensional vector space over $\Bbb F_2$, or the general linear group of a 2-dimensional vector space on $\Bbb F_4$? –  Olivier Bégassat Feb 14 '13 at 20:54
    
So to you $GL(n^2,q)$ stands for the general linear group of an $n$ dimensional vector space over $\Bbb F_q$? In that case, I think you should clarify this in your question, because your notation isn't standard and has mislead @DerekHolt and myself. Also, this greatly simplifies the task of finding the conjugacy classes! –  Olivier Bégassat Feb 14 '13 at 20:57
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Yes, please clarify, $\mathrm{GL}(4,2)$ usually means $4 \times 4$ invertible matrices over the field with $2$ elements. –  Andreas Caranti Feb 14 '13 at 22:42

2 Answers 2

up vote 4 down vote accepted

$GL_2(\mathbb{F}_2)$ has order six. The identity is always its own conjugacy class, so since the size of each conjugacy class divides the order of the group, the other two classes must be of order $2$ and $3$. Conjugate elements have the same order, and every group of even order contains an odd number of elements of order $2$, so we know the size $3$ class corresponds to elements of order $2$ and the size $2$ class corresponds to elements of order $3$. These are easy enough to calculate.

$$\text{Order }3: \left( \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right) , \left( \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right)$$ $$\text{Order }2: \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) , \left( \begin{array}{ccc} 1 & 0 \\ 1 & 1 \end{array} \right) , \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right)$$

So, along with the identity, these are the conjugacy classes of $GL_2(\mathbb{F}_2)$.


EDIT: Okay, so evidently OP wanted $\operatorname{GL}_4(\mathbb{F}_2)$ after all. This is a much more difficult question, especially because the field has even characteristic. It has been studied, however, and you can find a paper about it here that includes a table of the conjugacy classes with singer polynomials and explanations behind why everything works the way it does.

The conjugacy classes end up coming out like this: $$\begin{array}{rl}\text{Size} & \text{Order}\\ 1 & 1 \\ 105 & 2 \\ 210 & 2 \\ 112 & 3 \\ 1120 & 3 \\ 1260 & 4 \\ 2520 & 4 \\ 1344 & 5 \\ 1680 & 6 \\ 3360 & 6 \\ 2880 & 7 \\ 2880 & 7 \\ 1344 & 15 \\ 1344 & 15 \end{array}$$

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Fixed it for you. Be more careful next time :-). –  Willie Wong Feb 15 '13 at 8:41
    
I'm sorry, I've misled you. My original intention was to $4×4$ matrices, not $2\times2$. –  ofer Feb 15 '13 at 11:15
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@ofer I see. I will update my answer. –  Alexander Gruber Feb 16 '13 at 0:49

There is not much to add to the article by Gordon, Jarvis, and Shaw that Alexander Gruber linked to, it explains things well and in much detail. I would just like to show how you can get the list of conjugacy class sizes for a large part quite easily, getting much mileage out of a few general ideas. I'll give references in brackets to relevant results in that paper. One point I won't go into at all is the isomorphism $GL(4,\Bbb F_2)\cong A_8$ that Derek Holt mentioned, and which provides an entirely independent approach.

The main point will be classifying conjugacy classes of matrices $A\in GL(4,\Bbb F_2)$ first by the characteristic polynomial $\chi_A$, and if necessary by their minimal polynomial $\mu_A$ and other invariant factors. I will order the invariant factors opposite to what is usual (and what the paper does), namely with each next polynomial a divisor (rather than multiple) of the previous one (this has the advantage that the multiplicity of each irreducible factor is weakly decreasing along the list, and forms a partition of its multiplicity in $\chi_A$). Thus the first invariant factor is $\mu_A$, and the product of all invariant factors is $\chi_A$ [theorem 1.1]. Quite often (for instance when $\chi_A$ is square-free) there is just one invariant factor, and the minimal and characteristic polynomials coincide; I'll call this the regular case (as in regular element of a Lie group), although the paper calls uses the term "cyclic" [theorem 1.3(iii)] which is quite justified.

Since $\chi_A$ for $A\in GL(4,\Bbb F_2)$ can be any monic polynomial of degree $4$ over $\Bbb F_2$ with non-zero constant term, it is useful to list the irreducible polynomials up to that degree, excluding $X$. Then there remains one irreducible polynomial each of degrees $1,2$, which I shall call $P_1=X+1$ and $P_2=X^2+X+1$, two irreducible polynomials of degree $3$ namely $P_{3,1}=X^3+X+1$ and $P_{3,2}=X^3+X^2+1$, and three irreducible polynomials of degree $4$ namely $P_{4,1}=X^4+X+1$, $P_{4,2}=X^4+X^3+1$ and $P_{4,3}=X^4+X^3+X^2+X+1$ (in fact only the number of irreducibles in each degree really matters).

Computing the centraliser of any matrix $A$ will be the main task. Each matrix that commutes with $A$ must stabilise any kernel or image of any polynomial in $A$, and then of course also intersections and sums of those. In particular, the canonical decomposition of the space into a direct sum of spaces annihilated by mutually coprime factors of the minimal polynomial (each a power of an irreducible polynomial) [theorem 1.2] must be respected by every matrix commuting with $A$, and this leads to a product decomposition of the centraliser of $A$.

One very useful fact is that in the regular case, the only matrices that commute with $A$ are those in the algebra $F[A]$ of polynomials in $A$ [theorem 1.3(iii)]. This is because in this case there exists a "cyclic vector" $v$, namely one from which any vector can be obtained by applying a polynomial in $A$; then if $B$ commutes with $A$, one easily shows for $P\in F[A]$ with $P\cdot v=B\cdot v$ that $P=B$.

Another relatively easy case is when $A$ is semi-simple, i.e., diagonalisable over an extension field of $F$ (note that the only invertible matrices that are diagonalisable over $F=\Bbb F_2$ itself are the identity matrices!). Here the minimal polynomial is square-free, and the minimal polynomial of each component of the primary decomposition is an irreducible polynomial $P$; that component is a module over $F[X]/(P)$ which is a field, and the corresponding component of the centraliser is the appropriate general linear group over that field.

Now we can start our classification, after remarking that $\#GL(4,\Bbb F_2)=15\times14\times12\times7=20160$.

  1. $\chi_A$ is irreducible (three cases $\chi_A=P_{4,i}$ for $i=1,2,3$); then we are both in the regular and the semi-simple case. We have $F[A]\cong\Bbb F_{16}$ and the centraliser of $A$ in $GL(4,\Bbb F_2)$ is $C(A)=F[A]\setminus\{0\}\cong GL(1,\Bbb F_{16})=\Bbb F_{16}^\times$ which has $15$ elements. Each conjugacy class has size $20160/15=1344$, for a total of $3\times1344=4032$.

  2. $\chi_A=P_{3,i}P_1$ for $i=1,2$ (two cases). This is also a regular semi-simple case, with $F[A]\cong\Bbb F_8\times\Bbb F_2$ and $C(A)\cong\Bbb F_8^\times\times\Bbb F_2^\times$ which has $7\times1=7$ elements, and each conjugacy class has size $20160/7=2880$, for a total of $2\times2880=5760$.

  3. $\chi_A=\mu_A=P_2^2$, a single, regular case. One has $F[A]\cong F[X]/(P_2^2)$, and of the $16$ elements of this ring all but the $4$ ones divisible by $P_2$ are invertible, so $\#C(A)=12$, the conjugacy class has size $20160/12=1680$.

  4. $\chi_A=P_2^2$ and $\mu_A=P_2$, a single, semi-simple case. One has $C(A)\cong GL(2,\Bbb F_4)$ which has $15\times12=180$ elements; the conjugacy class has size $20160/180=112$ elements.

  5. $\chi_A=\mu_A=P_2P_1^2$, a single, regular case. One has $F[A]\cong \Bbb F_4\times F[X]/P_1^2$, and $C(A)=F[A]^\times$ has $3\times(4-2)=6$ elements, so the conjugacy class has $20160/6=3360$ elements.

  6. $\chi_A=P_2P_1^2$ and $\mu_A=P_2P_1$, a single, semi-simple case. One has $C(a)\cong \Bbb F_4^\times \times GL(2,\Bbb F_2)$ which has $3\times(3\times2)=18$ elements, and the conjugacy class has size $20160/18=1120$.

  7. $\chi_A=\mu_A=P_1^4$, a single, regular case. One has $F[A]\cong F[X]/(P_1^4)$ and $C(A)\cong F[A]^\times$ has $16-8=8$ elements, so the conjugacy class has size $20160/8=2520$.

  8. $\chi_A=P_1^4$ and $\mu_A=P_1^3$, with a second invariant factor $P_1$. This is a single case that is neither regular nor semi-simple, so we need to work a bit harder. Here $A$ is a unipotent matrix with Jordan blocks of size $3,1$; its centraliser is the same as that of $$ A-I=\begin{pmatrix}0&1&0&0\\0&0&1&0&\\0&0&0&0\\0&0&0&0\end{pmatrix} $$ which can be explicitly computed to have $1^2\times2^4=16$ elements. The conjugacy class has $20160/16=1260$ elements.

  9. $\chi_A=P_1^4$ and $\mu_A=P_1^2$, with a second invariant factor $P_1^2$. Again a case that is neither regular nor semi-simple; $A$ is a unipotent matrix with Jordan blocks of size $2,2$. Its centraliser is the same as that of $$ A-I=\begin{pmatrix}0&0&1&0\\0&0&0&1&\\0&0&0&0\\0&0&0&0\end{pmatrix} $$ which can be explicitly computed to to be a semidirect product of the vector space $\Bbb F_2^4$ with $GL(2,\Bbb F_2)$ which has $2^4\times(3\times2)=96$ elements. The conjugacy class has $20160/96=210$ elements.

  10. $\chi_A=P_1^4$ and $\mu_A=P_1^2$, with two more invariant factors $P_1$. Here $A$ is a unipotent matrix with Jordan blocks of size $2,1,1$; its centraliser is the same as that of $$ A-I=\begin{pmatrix}0&0&0&1\\0&0&0&0&\\0&0&0&0\\0&0&0&0\end{pmatrix} $$ which can be explicitly computed have $2^5\times(3\times2)=192$ elements. The conjugacy class has $20160/192=105$ elements. A sigh of relief, we needed at least one non-identity class with an odd number of elements!

  11. $\chi_A=P_1^4$ and $\mu_A=P_1$. Another semi-simple case, indeed the identity matrix. A singleton class.

A final check: $3\times1334+2\times2880+1680+112+3360+1120+2520+1260+210+105+1=20160$, Oof!

I did not mention the order of the elements in each class, but these depend only on the minimal polynomial, as described in this answer.

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