# A Structure of modulo-integer matrix group GL(r,Z(n))?

#### lpetrich

Over in the thread The eight-queens chess puzzle and variations of it | Physics Forums I discovered that with a toroidal board, one with periodic boundary conditions, the amount of symmetries becomes surprisingly large (A group-based search for solutions of the n-queens problem - ScienceDirect).

I will use a r-dimensional board with n squares along each dimension. A standard chessboard has r = 2 and n = 8. The toroidal-board symmetry group is an affine-symmetry group, a group whose elements are rotation/reflection and translation (R,T), elements that act on a vector x to make vector x' with $x' = R\cdot x + T$, where x, x', and T are r-vectors, R is an r*r matrix, and all components are in Zn. Affine symmetry thus generalizes the Euclidean group to components in arbitrary rings, like Zn under addition and multiplication modulo n. The rings do need additive and multiplicative identities, 0 and 1, for constructing the identity element R = identity matrix and T = zero vector.

For components being in a commutative ring, one can find the inverse of each R by using Cramer's rule. Though impractical for all but the smallest matrices, Cramer's rule is good for theoretical results. It is evident from that algorithm that invertibility requires that each R have a determinant that is a "unit" in that ring, an element with a multiplicative inverse. For $Z_n$, this is all elements relatively prime to n, or $Z^\times_n$. It is easy to show that the units of a ring form a group under multiplication.

So I must find the structure of the group of (R,T) for all invertible R and all T over Zn and number of dimensions r.

The first step is to note that with I the identity matrix, the affine group's subgroup of all (I,T) is a normal subgroup. Its structure is $(Z_n)^r$, (the component ring's additive group)r. Its quotient group is the group of R's under matrix multiplication, the homogeneous affine group. This group is $GL(r,Z_n)$, the general linear group over ring $Z_n$.

One can proceed further by selecting out all the elements with determinant 1, giving the special linear group: $SL(r,Z_n)$, The center of the general linear group is all elements $mI$ where m is in $Z^\times_n$, all the component ring's units times the identity matrix. The quotient group of $GL(r,Z_n)$ and its center is the projective general linear group $PGL(r,Z_n)$. The special linear group has a similar center, but with the further condition $m^r = 1$. Its quotient group is the projective special linear group $PSL(r,Z_n)$.

For r = 1, the group $GL(1,Z_n) = Z^\times_n$, the units group, and the groups $SL(1,Z_n)$, $PGL(1,Z_n)$, and $PSL(1,Z_n)$ are all the identity group.

For n = 2, $Z^\times_2$ is the identity group, and $GL(r,Z_2) = SL(r,Z_2) = PGL(r,Z_2) = PSL(r,Z_2)$.

One can simplify these groups in another direction, by using a theorem of matrices over rings and their ideals (Combinatorics of Vector Spaces over Finite Fields). A two-sided ideal I of a ring R satisfies $R\cdot I = I \cdot R = I$, with a left ideal and a right ideal being for only one side. An ideal of the integer ring $Z$ is $n \cdot Z$ for integer n.

The theorem: consider matrix ring $M(r,C)$ of r*r matrices over component ring C. Consider ideals of C, Jk, that are "disjoint". Their intersection $J = \cap_k J_k$. Then $M(r,C/J) = \prod_k M(r,C/J_k)$, where each C/J is a quotient ring. This result readily gives analogous results for the GL, SL, PGL, and PSL groups.

Applying to Zn, that ring is is the quotient ring Z/(nZ), and I've sometimes seen that notation used instead. An appropriate set of disjoint ideals is from prime factors: $n = \prod_p p^m$, powers m of prime factors p. Thus, $Z_n = \prod_p Z_{p^m}$. This is also true of the multiplicative group over its units, $Z^\times_n = \prod_p Z^\times_{p^m}$, of its matrix rings $M(r,Z_n) = \prod_p M(r,Z_{p^m})$, and of the GL, SL, PGL, and PSL groups.

Multiplicative group of integers modulo n - Wikipedia goes further with $Z^\times_n = \prod_p Z^\times_{p^m}$, stating the structure of each group $Z^\times_{p^m}$:
• For p = 2, $Z^\times_2$ is the identity group and $Z^\times_{2^m} = Z_{2^{m-2}} \times Z_2$.
• For odd p, every other value: $Z^\times_{p^m} = Z_{p^{m-1}} \times Z_{p-1} = Z_{(p-1)p^{m-1}}$.

Continuing with $M(r,Z_{p^m})$ and its groups, I've only seen stuff on the groups for a related matrix ring, $M(r,GF(p^m))$, with its component ring being a Galois field. But $Z_{p^m}$ is only a field for m = 1: GF(p).

However, that combinatorics reference states the order of $GL(r,Z_{p^m})$. It is $p^{(m-1)r^2} \prod_{k=0}^{r-1} (p^r - p^k)$ or $p^{(m-1)r^2} |GL(r,Z_p)|$. More generally,
$$|GL(r,Z_n)| = n^{r^2} \prod_p \prod_{k=1}^r \left( 1 - \frac{1}{p^k} \right)$$
over all prime factors p of n. For r = 1, one gets the formula for the Euler totient function. But for Galois fields,
$$|GL(r,GF(q))| = q^{r^2} \prod_{k=1}^r \left( 1 - \frac{1}{q^k} \right)$$

My question: What has been done on the structure of the group $PSL(r,Z_{p^m})$ for r > 1 and m > 1? For r = 1, it is the identity group. For m = 1, it is simple for all r and m except for r = 2 and m = 2 and 3. That is as far as I can go without doing brute-force calculations.

Related Linear and Abstract Algebra News on Phys.org

#### lpetrich

I found a site with some helpful theorems: Groupprops: the Group-Properties Wiki Some of its theorems are useful for this problem, because Z(n) is a commutative unital ring (ring with unity), even though it is not a field when n is composite.

I first consider the relation between Comm(GL(r,R)), the commutator subgroup of GL(r,R) and SL(r,R): Derived subgroup of general linear group is special linear group - Groupprops -- true in all but some special cases.

The first part of the proof consists of finding the determinants of the elements of Comm(GL(r,R)). They are all 1, and thus, Comm(GL(r,R)) is a subgroup of SL(r,R). This does not complete the proof, because the commutator subgroup may be a proper subset of the special linear group. So we consider whether the subgroup relation goes in the other direction.

For the second part, we need results Every elementary matrix of the first kind is a commutator of invertible matrices - Groupprops and Elementary matrices of the first kind generate the special linear group over a field - Groupprops.

For matrix size r >= 3, the first result follows from Every elementary matrix of the first kind is a commutator of elementary matrices of the first kind - Groupprops. For r = 2, we consider
$$\text{Comm} ( \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} , \begin{pmatrix} 1 & b c/(a-b) \\ 0 & 1 \end{pmatrix} ) = \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}$$
The first matrix is invertible if a and b are units of R, elements with reciprocals (multiplicative inverses). The second matrix is also possible only if (a-b) is also a unit of R. Specializing to Z(n), this decomposition is always possible for n odd, but not for n even.

For the second result, "The proof relies on a slight variation of the usual Gauss-Jordan elimination procedure. The variation is necessary because we do not have access to the other types of elementary matrices (permutations and scalar multiplications)." No further details, though one can include even permutations because GL(r,R) has all permutation matrices for size r and its commutator all even-permutation matrices for size r. With permutations, it is straightforward to work out how G-J elimination works for each column. For r = 2, the only non-identity permutation is the interchange one, an odd one, and that makes it more difficult. The algorithm is

Starting with {a1, a2, ..., a(r)} permute it so that the element with smallest nonzero absolute value will be first. Now calculate {a1, a2-m2*a1, ... a(r)-m(r)*a1} such that each a(k)-m(k)*a1 has absolute value less than a1. Repeat until only one nonzero value remains. Since the matrix must be invertible, this remaining value must be a unit in R.

So GL(r,Z(n)) has commutator group SL(r,Z(n)) except if r = 2 and n is even.

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#### lpetrich

Turning to the special linear group SL(r,R), I use the results in Special linear group is perfect - Groupprops, A group G is "perfect" if Comm(G) = [G,G} = G. The proof parallels the proof for the general linear matrix's commutator, except that it uses Every elementary matrix of the first kind is a commutator of unimodular matrices - Groupprops -- unimodular instead of invertible, where "unimodular" means "determinant = 1". For matrix size r >= 3, we use the previously-mentioned result about how every elementary matrix is the commutator of two elementary matrices. For r = 2, we consider
$$\text{Comm} ( \begin{pmatrix} a/b & 0 \\ 0 & b/a \end{pmatrix} , \begin{pmatrix} 1 & b^2 c/(a^2-b^2) \\ 0 & 1 \end{pmatrix} ) = \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}$$
So all of a, b, and a^2-b^2 must be units in R. This is only possible if n is not evenly divisible by 2 or 3 inclusive.

So in summary:
• Comm(GL(r,Z(n))) = SL(r,Z(n)) if r >= 3 or else r = 2 and n is not divisible by 2
• Comm(SL(r,Z(n))) = SL(r,Z(n)) if r >= 3 or else r = 2 and n is not divisible by 2 or 3
Since products of different prime powers split apart, we need only consider each individual prime power. Taking the case of r = 2 and the cases of powers of 2 and 3, the groups' orders are:
• $|GL(2,Z(2^m))| = 2^{4m-3} \cdot 3$
• $|GL(2,Z(3^m))| = 2^3 \cdot 3^{4m-3}$
By Burnside's theorem, all these groups are solvable. This theorem states that all groups with order $p^a q^b$ are solvable, for primes p and q. The same is true for the related PGL, SL, and PSL groups. That means that all these groups have a commutator series that ends in the identify group, and thus that SL(2,Z(n)) is not perfect for n divisible by 2 or 3.

Though the groups SL(r,Z(n)) are perfect for r >= 3 or n not divisible by 2 or 3, that does not mean that their projective counterparts are simple. Consider PSL(2,Z25). Its order is 7500, but perusing List of finite simple groups - Wikipedia reveals no finite simple group with that order. So this group must have a non-central normal subgroup with a nonabelian quotient group.

Has anyone ever tried to find this structure?

#### lpetrich

I must examine the issue of the projective linear groups in more detail. These groups are quotient groups of the matrix groups with respect to their scalar-matrix subgroups, subgroups of multiples of the identity matrix. So let us consider quotient groups and commutators of them more generally.

Quotient groups may be interpreted as results of homomorphisms, group-to-group functions. Homomorphisms are always surjections and sometimes bijections. A homomorphism has a kernel, all the elements that map onto the identity element of the new group, and that is the divisor group for a quotient group. For group G with subgroup H and homomorphism F, F(H) is a subgroup of F(G), and kernel(F,H) is a subgroup of kernel(F,G).

Since I've been discussing commutation, I consider how that works with homomorphisms. Comm(F(G)) = F(Comm(G)) -- the two operations commute. So a quotient group for a commutator group is the commutator group of the original quotient group.

Thus, when Comm(GL(r,R)) = SL(r,R), Comm(PGL(r,R)) = PSL(r,R). Likewise, when Comm(SL(r,R)) = SL(r,R), then Comm(PSL(r,R)) = PSL(r,R).

This means that when PSL(r,R) is a perfect group, any quotient groups that it might have are also perfect groups, something true in general for perfect groups. So for PSL(r,Z(n)) for r >= 3, or else r = 2 and n not divisible by 2 or 3, all its quotient groups will be perfect, as it itself is.

#### fresh_42

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These groups are quotient groups of the matrix groups with respect to their scalar-matrix subgroups, subgroups of multiples of the identity matrix.
Indeed, and it makes me doubt $|\operatorname{PSL}(2,\mathbb{Z}_{25})| = 7,500\,.$ It looks to "smooth". E.g. I would have expected $\operatorname{PSL}(2,\mathbb{Z}_{25}) \cong A_1(25)$ which has $7,800$ elements if I made no mistake.

However, I haven't checked neither result - just thinking.

#### lpetrich

I calculated the numbers myself by brute force with Mathematica. I generated the appropriate matrices and counted them. Results:
• All 2*2 matrices over Z25: $|M(2,Z_{25})| = 390,625$
• All with determinant value relatively prime to 25: $|GL(2,Z_{25})| = 300,000$
• All with determinant value equal to 1: $|SL(2,Z_{25})| = 15,000$
• Number of scalar matrices in $GL(2,Z_{25})$: $20$
• Number of scalar matrices in $SL(2,Z_{25})$: $2$
• Quotient group $|PGL(2,Z_{25})| = 15,000$
• Quotient group $|PSL(2,Z_{25})| = 7,500$
• Multiplication group $Z^\times_{25} \sim Z_5 \times Z_4$
• Its order: $20$
• Product of gcd(cycle lengths, 2): $2$
All these results agree with the formulas that I had earlier presented.
• All 2:2 matrices over GF(25): $|M(2,GF_{25})| = 390,625$
• All with nonzero determinant value: $|GL(2,GF_{25})| = 374,400$
• All with determinant value 1: $|SL(2,GF_{25})| = 15,600$
• Number of scalar matrices in $GL(2,GF_{25})$: $24$
• Number of scalar matrices in $SL(2,GF_{25})$: $2$
• Quotient group $|PGL(2,GF_{25})| = 15,600$
• Quotient group $|PSL(2,GF_{25})| = 7,800$
• Multiplication group $GF_{25} \sim Z_{24}$
• Its order: $24$
• Product of gcd(cycle lengths, 2): $2$
All these results agree with well-known results for these groups.

Scalar matrices = multiples of the identity matrix.

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#### fresh_42

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The groups PSL(n, q) are simple unless n =2 and q = 2 or 3.
https://www.johndcook.com/blog/2018/09/08/projective-special-linear/

Finite fields
The projective special linear groups PSL(n, Fq) for a finite field Fq are often written as PSL(n, q) or Ln(q). They are finite simple groups whenever n is at least 2, with two exceptions: L2(2), which is isomorphic to S3, the symmetric group on 3 letters, and is solvable; and L2(3), which is isomorphic to A4, the alternating group on 4 letters, and is also solvable.
https://en.wikipedia.org/wiki/Projective_linear_group

https://groupprops.subwiki.org/wiki/Projective_special_linear_group:PSL(2,25)

Edit: For further reads, these papers might be of interest:
https://ac.els-cdn.com/002186936690007X/1-s2.0-002186936690007X-main.pdf?_tid=35d0d7b9-770f-4329-ba41-926e0d53ddb6&acdnat=1552658880_af4130308e2413240261f1f23664fd1f

or generally search for finite Chevalley groups.

#### lpetrich

Thanx for your references, but even after reading them, I still can't figure out how to go from Lie-algebra commutators to Chevalley-group elements.

Perfect group - Wikipedia and Perfect group - Groupprops -- some nice stuff in it, like how a perfect group's quotient groups are all perfect, and how a perfect group's quotient group for its center is a group with an identity-group center -- the "upper central series" of quotienting out group centers stops there.

The images of commutator subgroup under homomorphisms, and images of solvable groups - Mathematics Stack Exchange states what I'd stated earlier, that Comm(F(G)) = F(Comm(G)) where F is a homomorphism of group G.

I will now consider other sorts of matrix groups, the orthogonal, symplectic, and unitary ones. The orthogonal ones are O(r,R) for matrix size r and commutative ring R, and for determinant = 1, SO(r,R). Matrices M satisfy
$$MM^T = M^TM = I$$
The symplectic ones Sp(2r,R) have an antisymmetric two-form, J, that satisfies $J^2 = - I$. Matrices M satisfy
$$MJM^T = J$$
For unitary ones, we need to define an element conjugation function C(R) -> R such that C(a+b) = C(a) + C(b) and C(a*b) = C(a)*C(b) (distributive property), and C(C(a)) = a (involution property). For complex numbers, C can be complex conjugation, while for finite fields $GF(q^2)$, one uses $C(x) = x^q$. Matrices M are given by
$$M C(M^T) = C(M^T) M = I$$
The combination $C(M^T)$ is the ring's counterpart of the Hermitian conjugate.

#### fresh_42

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Comm(F(G)) = F(Comm(G))
This is obvious. $F([g,h])=F(g^{-1}h^{-1}gh)=F(g^{-1})F(h^{-1})F(g)F(h)=F(g)^{-1}F(h)^{-1}F(g)F(h)=[F(g),F(h)]\,.$
Thanx for your references, but even after reading them, I still can't figure out how to go from Lie-algebra commutators to Chevalley-group elements.
There is no direct correspondence on the level of elements, not even on group level. Nevertheless there is a correspondence. I haven't studied any details here, but if you look at the list of simple Lie algebras and the list of finite simple groups, you see the strong relation. It is given by the geometry of root spaces, aka Dynkin diagrams.

See: https://en.wikipedia.org/wiki/Group_of_Lie_type or for a general list of vocabulary https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/
and if you aren't already a regular reader of Terence Tao's blog (which I recommend):
https://terrytao.wordpress.com/tag/chevalley-group/

#### lpetrich

I first note that Sp(2r,R) has even matrix size, and that a common statement of conserved two-form matrix J in it is
$$J = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$$
The orthogonal case has |O(r,R)| = |SO(r,R)| * (number of square roots of unity in R). For real numbers, that number of square roots is 2, while this is not necessarily the case for other rings, like Z(8). Its units are {1, 3, 5, 7}, and the squares of them in the ring are all 1. Thus, Z(8) has four square roots of unity.

The center of GL(r,R) is (units of R) * I, the center of SL(r,R) is (rth roots of unity in R) * I, something that can easily be proved with matrices equal to the identity matrix with one non-diagonal element, matrices in SL(r,R) like
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$
However, O(r,R) an SO(r,R) do not contain such matrices, and the most that one can say in general about these groups' centers is that the center of O(r,R) contains (2r'th roots of unity in R) * I, and the center of SO(r,R) contain (rth roots of unity in R) * I.

For r at least 3, we can consider matrices like these ones in SO(3,R):
$$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} ,\ \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Thus, if R's characteristic is not 2, that is, 1 + 1 is nonzero in R, the centers of O(r,R) and SO(r,R) are multiplies of I.

#### lpetrich

For characteristic 2 (2 -> 0, -1 -> 1, etc.), it is more tricky. In general, the matrix that commutes with the above two matrices is
$$\begin{pmatrix} a & b & b \\ b & a & b \\ b & b & a \end{pmatrix}$$
where $a^2 = 1$ and $b^2 = 0$. For GF(2^n), the only solution is the identity matrix, while one can construct a ring with both characteristic 2 and nonzero square roots of zero, a ring like $\{0, 1, z, 1+z\}$ where $z^2 = 0$. The fourth element in this one is a square root of unity. For O(3,{0,1,z,1+z}), the center is a = 1 or 1+z and b = 0 or z, while SO(3,{0,1,z,1+z}) has a = 1 and both values of b.

I must note that the orthogonality condition means that the multipliers of the scalar (multiple of identity) elements of the groups' centers must be square roots of unity. This means that the center of O(r,R) includes {a*I} for square roots of unity a, and that the center of SO(r,R) includes all these a's that are also rth roots of unity.

It is well-known that elements of SO(3,R) can be constructed from elements of the unit quaternion group UQ(R) for ring R, and also that elements of SO(4,R) can be constructed from pairs of UQ(R) elements. The quaternion elements cover the SO groups for the real numbers, giving homomorphisms UQ(R) -> SO(3,R) with kernel {id,-id}, and UQ(R)*UQ(R) -> SO(4,R) with kernel {(id,id), -(id,id)} for quaternion identity id.

But that is not necessarily the case for rings like Z(n) and GF(p^n).

In particular, for characteristic 2, the unit-quaternion group becomes abelian, the quaternion elements in SO(3,R) reduce to the identity group, and the quaternion-product elements in SO(4,R) reduce to a linear function of the product of those two quaternions.

For Z2, I find: UQ(Z2) `= (3 0's, 0 1's) and (1 0, 3 1's) (order 8), SO(3,Z2) = permutations of the 3*3 identity matrix (order 6), and SO(4,Z2) = permutations of the 4*4 identity matrix with (1 - those elements) added (order 48).

So it's SO(3): (1 out of 6) and SO(4): (8 out of 48).

For Z3, I find UQ: 24, SO(3): (12 out of 24), and SO(4): (288 out of 576).

#### lpetrich

Turning to r = 2, O(2,R) and SO(2,R) have element matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

Orthonormality gives constraints $a^2 + b^2 = 1$, $a^2 = d^2$, $b^2 = c^2$, $ac + bd = 0$, $ab + cd = 0$, and determinant $ad - bc$. For a field, it is easy to find the possible solutions: $d = a, c = -b$ (det = 1) and $d = -a, c = b$ (det = -1):
$$\begin{pmatrix} a & b \\ -b & a \end{pmatrix} (\det = 1) , \begin{pmatrix} a & b \\ b & -a \end{pmatrix} (\det = -1)$$
But a more general ring has the complication of zero divisors and possible nonzero square roots of zero. In Z6 has zero divisors 2, 3, and 4, and in Z4, 2 is a square root of zero. They add additional matrices to the field-value matrices, matrices whose determinants may be additional square roots of unity. For instance, Z8 has such additional ones as {{1,0},{0,3}}, with det = 3, and {{1, 2}, {6, 3}}, with det = 7 (-1 mod 8).

But SO(2,R) has an additional equation, the determinant equation $ad - bc = 1$. That offers some simplifications. Multiply that equation by d and $ab + cd = 0$ by b, add, and one gets $ad^2 + ab^2 = d$, With $a^2 + b^2 = 1$, $a^2 = d^2$, and $b^2 = c^2$, this simplifies into $d = a$. Likewise, one finds that $c = -b$. This gives the (det = 1) matrices for the field case.

The SO(2,R) matrices all commute with each other, making SO(2,R) abelian, though not necessarily cyclic.

Their parameters have this composition law, for parameter sets $(a_1, b_1), (a_2, b_2)$ giving $(a_{12}, b_{12})$:
$$a_{12} = a_1 a_2 - b_1 b_2 ,\ b_{12} = a_1 b_2 + b_1 a_2$$

#### lpetrich

While SO(2,R) is abelian, O(2,R) is nonabelian in most cases, with a quotient group for SO(2,R) that is the group of square roots of unity in R. This can be more than {1,-1}. For instance, in Z8, it is {1,3,5,7}, the last element being -1 mod 8.

There are some well-known interrelationships between small Lie algebras, distinguished here from the groups with small letters, and these can guide us in searching for interrelationships between the matrix groups:
• so(2) ~ u(1)
• so(3) ~ su(2) ~ sp(2)
• so(4) ~ su(2) * su(2)
• so(5) ~ sp(4)
• so(6) ~ su(4)
Though it is the spinor versions that most directly correspond. The vector representations SO(n) are derived from the spinor representations Spin(n) by finding product representations. Thus,
• SU(2) ~ Sp(2) ~ Spin(3) ... SO(3) ~ square of Spin(3)
• SU(2) ~ each part of Spin(4) ... SO(4) ~ product of both parts of Spin(4)
• Sp(4) ~ Spin(5) ... SO(5) ~ square of Spin(5)
• SU(4) ~ each part of Spin(6) ... SO(6) ~ square of one part of Spin(6)
SU(n) corresponds to SL(n,R) here.

The representation breakdowns:
• so(3): (2 spinor)^2 = (3 vector) + (1 scalar)
• so(4): ((2,1) spinor) * ((1,2) spinor) = ((2,2) vector)
• so(5): (4 spinor)^2 = (10 adjoint) + (5 vector) + (1 scalar)
• so(6): (4 spinor)^2 = (10) + (6 vector)

I find: Sp(2,R) ~ SL(2,R) (matrices equal)

However, the others may have only imperfect correspondence, as with the quaternions. The spinor squares or products may give only part of the complete group.

#### lpetrich

Thanx. It mentioned Sp(2,complexes) ~ SL(2,complexes) and SO(3,reals) ~ SU(2,complexes). I'd earlier proved that Sp(2,R) = SL(2,R) for *any* commutative unital ring R.

For real-valued matrices, we have constructions
• SO(2,1,reals) ~ square of SL(2,reals)
• SO(2,2,reals) ~ SL(2,reals) * SL(2,reals)
• SO(3,2,reals) ~ square of Sp(4,reals)
• SO(3,3,reals) ~ square of SL(4,reals)
The squares can be broken down by symmetry:
• (2) ^ 2 = S: (3 vector), A: (1 scalar)
• (4)^2 = S: (10 adjoint), A: (5 vector) + (1 scalar)
• (4)^2 = S: (10), A: (6 vector)
S = symmetric, A = antisymmetric

Generalizing to arbitrary c.u.r.'s R, the construction of SO(2,1,R) from SL(2,R) will be a subgroup of SO(2,1,R), even if a proper subgroup in many cases. Likewise, this construction applied to GL(2,R) will give a subgroup of O(2,1,R).

This construction applied to GL(2,Z2) ~ SL(2,Z2) gives O(3,Z2) ~ SO(3,Z2), but for Z3, I find:
• |O(3,Z3)| = 48
• |SO(3,Z3)| = 24
• |O(3,Z3) from GL(2,Z3)| = 24 (12 each for det = 1, 2)
• |SO(3,Z3) from SL(2,Z3)|= 12

This is likely the case for the other constructions.

#### lpetrich

I recently found a general expression for O(2,R) matrices for commutative unital rings R. It involves square roots of unity, all elements s of R that satisfy $s^2 = 1$. They form a group under multiplication, and this group has form $(Z_2)^n$ with order $2^n$ for some n. The square roots of unity contain 1 (of course!) and also -1, but for characteristic equal to 2, -1 = 1 and is thus not distinct from 1. Here is the general expression that I've found:
$$\begin{pmatrix} a & b \\ - s b & s a \end{pmatrix}$$
where $a^2 + b^2 = 1$ and s, a square root of unity, is the value of the matrix's determinant. If the matrix elements are either 0 or square roots of unity, then the possible O(2,R) ones are
$$\begin{pmatrix} s_1 & 0 \\ 0 & s_2 \end{pmatrix} ,\ \begin{pmatrix} 0 & s_1 \\ s_2 & 0 \end{pmatrix}$$
and the possible SO(2,R) ones are
$$\begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} ,\ \begin{pmatrix} 0 & s \\ - s & 0 \end{pmatrix}$$
This subgroup of O(2,R) has order $2^{2n+1}$, and this subgroup of SO(2,R) has order $2^{n+1}$.

#### lpetrich

Returning to Chevalley groups, I gather that they are associated with the "universal covering algebras" of simple Lie algebras, where the algebras' generators have no properties other than commutation: $[L_i,L_j] = f_{ij}{}^k L_k$ for structure constants f. The logical sort of group is the algebra's automorphism group, with group matrices given by $L_i' = A_i{}^j L_j$, and with them satisfying $A_i{}^a A_j{}^b f_{ab}{}^c = f_{ij}{}^k A_k{}^c$.

Claude Chevalley discovered that it is always possible to put a semisimple Lie algebra into a basis where the f's are all integers. This means that the A's can be matrices over arbitrary fields, including finite ones.

Automorphisms are divided into inner ones, generated by the group or algebra itself, and outer ones, all the others. An inner one with a small departure from identity is $$A_i{}^j = \delta_i^j + f_{ik}{}^j a^k + O(a^2)$$ for some vector a, and more generally, $A = \exp(f\cdot a)$. This is a plausible definition only for Cauchy-closed characteristic-0 fields like real numbers and complex numbers, and not for finite fields.

#### lpetrich

There are additional groups of Lie type: the "twisted" Steinberg ones. These ones are for algebras whose Dynkin diagrams have nontrivial symmetries, and for fields with the same kinds of symmetries.

The Dynkin diagrams for A(n) (n >=2), D(n), and E6 both have twofold symmetry, symmetry under flipping.

The Dynkin diagram for D4 has threefold symmetry, symmetry under three flips and two order-3 rotations.

For the twofold-symmetry algebras, we need twofold symmetry for the field. This is provided by a flipping operator, a degree-2 automorphism or involution. For complex numbers, the appropriate flipping operator is complex conjugation, while for finite field GF(q^2), the appropriate operator is x -> x^q. So to the automorphism matrices we add a field-flip operator.

The algebra D4 has an additional kind of symmetry, threefold rotation, and for finite field GF(q^3), the appropriate operator is x->x^q. As earlier, we add this operator o the automorphism matrices.

So we have Chevalley groups A(n,q), B(n,q), C(n,q), D(n,q), G2(q), F4(q), E6(q), E7(q), and E8(q) and Steinberg groups 2A(n,q^2), 2D(n,q^2), 2E6(q^2), and 3D4(q^3).

Another sort of additional group has q = 2^(2m+1). This gives the Ree groups 2B2(q), 2F4(q), and 2G2(q). I can't find any simple explanation of how they are constructed, however. But Chevalley, Steinberg, and Ree groups cover all the groups of Lie type. The only remaining finite simple groups are the prime-order cyclic ones, the alternating ones with parameter at least 5, and the 26 sporadic groups.

"Structure of modulo-integer matrix group GL(r,Z(n))?"

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