# Class equation of subgroup of $SL(4,\mathbb{F}_2)$

Can you point me toward a computation-light derivation of the class equation of the subgroup of $SL(4,\mathbb{F}_2)$ consisting of upper-triangular matrices with 1's on the main diagonal?

The question is motivated by a homework problem from a class in representation theory. The homework problem is to describe all the irreducible complex representations of the above group. I began by doing pages of calculations in the group, trying to find the conjugacy classes by hand, so I would know how many representations to expect and might be able to play with the numbers to find their dimensions. I turned away from this because it was getting too computationally heavy for me to feel safe I wasn't making any small mistakes. I've started looking for representations in an ad-hoc way, but I don't know how many to expect or what dimensions, so this is also pretty hairy. Also, I'd like to organize my results using the character table, but for this again I would need to know the group's decomposition into conjugacy classes.

Besides specific computations, here is what I have figured out about the group: it is a semidirect product of $D_4$ and $C_2\times C_2\times C_2$. This is because it acts on vectors in $\mathbb{F}_2^4$ with last component 0. The action coincides with the action of the upper left $3\times 3$ block on $\mathbb{F}_2^3$, so this gives us a homomorphism onto the subgroup of $SL(3,\mathbb{F}_2)$ of upper-triangular matrices with 1's on the main diagonal. This latter group is isomorphic to $D_4$, and the restriction of the homomorphism to the subgroup of the original group which is 0 off the diagonal in the last column is an isomorphism. The kernel is the subgroup of matrices that are zero off the main diagonal away from the last column, and this is isomorphic to $C_2\times C_2\times C_2$. However I am unsure how to use this information to find the class equation.

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You calculations should become simpler if you consider the semidirect product $$G := \left(\begin{array}{cccc} 1 & * & * & *\\ & 1 & * & *\\ & & 1 & *\\ & & & 1 \end{array}\right) = \left(\begin{array}{cccc} 1 & & * & *\\ & 1 & * & *\\ & & 1 & \\ & & & 1 \end{array}\right)\rtimes \left(\begin{array}{cccc} 1 & * & & \\ & 1 & & \\ & & 1 & *\\ & & & 1 \end{array}\right) =: N\rtimes H.$$

Both $N$ and $H$ are elementary abelian, and you have mostly to understand the conjugation action of the three nonzero elements of $H$ on $N$.

EDIT: As you finished your assignment I now post an almost complete solution.

Denoting the elements of $H$ as $1, x, y$ and $z$ we get: $$x^{-1}\cdot\left(\begin{array}{cc} b & a \\ d & c \end{array}\right)\cdot x := \left(\begin{array}{cccc} 1 & 1 & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{array}\right)\cdot\left(\begin{array}{cccc} 1 & & b & a\\ & 1 & d & c\\ & & 1 & \\ & & & 1 \end{array}\right)\cdot\left(\begin{array}{cccc} 1 & 1 & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{array}\right) = \left(\begin{array}{cccc} 1 & & b+d & a+c\\ & 1 & d & c\\ & & 1 & \\ & & & 1 \end{array}\right) = \left(\begin{array}{cc} b+d & a+c\\ d & c \end{array}\right)$$

and (only in short notation) $y^{-1}\cdot\left(\begin{array}{cc} b & a \\ d & c \end{array}\right)\cdot y = \left(\begin{array}{cc} b & a+b\\ d & c+d \end{array}\right)$

$N$ being normal and $H$ being abelian, the conjugacy classes are contained in the cosets $hN$ for $h \in H$. As $N$ is abelian, one can determine the centralizers of the elements of $N$ by looking at the action of $x$ and $y$ on $N$. One obtains that $N$ contributes $2\cdot 1 + 3\cdot 2 + 2\cdot 4$ to the class equation of $G$.

Now consider the centralizers/conjugacy classes of $X := \langle x, N\rangle$: $N$ is abelian and has index $2$ in $X$. By the action of $x$ on $N$ is $C := \left(\begin{array}{cc} * & *\\ 0 & 0 \end{array}\right) = \mathrm{Z}(X)$ the center of $X$, and $xC$ is the conjugacy class of $x$ in $X$. As $N$ is abelian, the elements $t$ of $X\setminus N$ all have centralizers in $X$ of order $4$ and their conjugacy classes (in $X$) are the cosets $tC$.

As $X$ has index $2$ in $G$ (i.e., is normal of prime index), for elements $t \in X\setminus N$ either the centralizer grows by factor $2$ or its conjugacy class is merged with another conjugacy class.

As the centralizer of $y$ in $G$ is $\langle x, \left(\begin{array}{cc} 0 & *\\ 0 & * \end{array}\right)\rangle$, there are two conjugacy classes (of $G$) of order $4$ contained in $xN$ ("growing centralizer"). As $y^{-1}\cdot x\left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)\cdot y = x\left(\begin{array}{cc} 0 & 0\\ 1 & 1 \end{array}\right)$ the other two conjugacy classes merge to a conjugacy class of $G$ of order $8$ contained in $xN$. Hence, $xN$ contributes $2\cdot 4 + 8$ to the class equation of $G$.

By symmetry, i.e., using automorphisms of $G$ fixing $N$ that map $x$ to $y$ (easy) rsp. $x$ to $z$ (not too difficult), $yN$ and $zN$ contribute the same as $xN$ does to the class equation, which is $$G = 2\cdot 1 + 3\cdot 2 + 2\cdot 4 + 3\cdot[2\cdot 4 + 8],$$ as Ben already commented.

Extra credit: Find the class equation for $\mathbb{F}_2$ generalized to any finite field $\mathbb{F}_q$.

(Hint: Conjugate with diagonal matrices)

EDIT 2: The action of the element $x$ on $N$ has a two-dimensional centralizer $C_N(x)$ (viewing the elementary abelian group as $\mathbb{F}_2$-vector space) that equals the commutator $[x, N]$. The same holds for $y$ and $z$, and the centralizers of any pair intersect one-dimensionally. As $N$ is elementary abelian, one can find an automorphism of $G$ exchanging $x$ and $z$ and fixing $N$ (not pointwise, but as a set):

One automorphism exchanging $x$ and $z$ maps a basis of $N$ like this:

$\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)\mapsto \left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right), \left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)\mapsto \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$, whereas $\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)$ and $\left(\begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right)$ are fixed.

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Additional hint: If two elements are conjugated in $G$, they are also conjugated mod $N$, which is abelian... ($N$ for example contributes $2\cdot 1 + 3\cdot 2 + 2\cdot 4$ to the class equation of $G$ as you easily deduce using the action of $H$ on $N$). –  ego Feb 24 '12 at 8:50
+1 This is all very helpful. You are saying that every conjugacy class is contained in a coset of $N$... –  Ben Blum-Smith Feb 24 '12 at 18:54
Using your hint I am getting $64=2\cdot 1 + 3 \cdot 2 + 2\cdot 4 + 3\cdot [2\cdot 4 + 8]$ as the class equation, do you agree? (Here I have organized the class equation into cosets of $N$.) –  Ben Blum-Smith Feb 25 '12 at 16:58
Hmmm... I am skeptical of your claim about there being an automorphism of $G$ fixing $N$ that maps $x$ to $z$. According to purely numerical considerations, there should be 2 faithful 4-dimensional irreducible representations of $G$. Since $x,y,z$ are not conjugate in $G$, any automorphisms sending $x$ to $y$ or $z$ should be outer. But if they exist, then $|Aut(G)|\geq 3$, thus (by composing one representation with the automorphisms) there should be at least 3 irreducible 4-dimensional representations of $G$. Or are some of the representations obtained this way conjugate to each other? –  Ben Blum-Smith Feb 27 '12 at 23:25
... With more careful examination I found that the 4-dimensional representations are fixed (up to equivalence) by composition with these automorphisms, so that answers my doubt... –  Ben Blum-Smith Mar 1 '12 at 22:54