Group theory -- mathematics of symmetries

[lpetrich]

Returning to p-groups, one can generalize the argument that shows that an order-p² group is abelian. If a group with order pⁿ has a center that contains a group with order p^n-1, then it is abelian. This means that every nonabelian p-group has a center order p^m, where 1 <= m <= n-2. Thus, the smallest nonabelian prime power is 3.


Now for stacked quotient groups. Consider a group G with a normal subgroup H and quotient group Q = G/H. Every element in G can be interpreted as being tagged by some element of Q. To use more group-theoretic language, there is a homomorphism from G to Q. Now consider a normal subgroup of Q, J, and its quotient group R = Q/J. Every element of Q can be tagged with an element of R, and likewise every element of G can be tagged with an element of R.

The kernel of a homomorphism is the part of the input group that maps onto the identity of the output group. A kernel is always a normal subgroup. The kernel of G -> R will thus be a normal subgroup K, with G/K = R.

The direct product of two groups has both input groups as normal subgroups with the other one for each one being the quotient group. Thus if G = G1 * G2, then G2 = G/G1 and G1 = G/G2.

By the Chinese remainder theorem, the cyclic group Z(m*n) = Z(m)*Z if m and n are relatively prime. Thus, Z = product of Z(maximum power of a), going over every prime factor. In turn, Z(pⁿ) has subgroups Z(p^m), each one with quotient group Z(p^n-m).


Applied to p-groups, a p-group with order pⁿ has a normal subgroup with order p^n-1 with quotient group Z(p). Continuing, we find a sequence of normal subgroups that ends at the identity group, with each pair having quotient group Z(p). Thus, every p-group is solvable.

Going away from p-groups again, there is a theorem called Burnside's theorem that states that every group with order pⁿ*q^m for primes p and q is solvable. Another theorem, the Feit-Thompson or odd-order theorem, states that every group with odd order is solvable. Burnside's theorem is somewhat complicated to prove, and the odd-order theorem even worse. For identifying all the finite simple groups, the proof is much, much worse, running over 10,000 journal pages.


So I'll stick to simpler stuff.

 

[lpetrich]

I'll now consider the smallest groups. These theorems save us a lot of work in finding them.

For order 2, the only one is Z2, since 2 is a prime number.

Likewise, for order 3, the only one is Z3.

For order 4, we note that 4 = 2², meaning that it is abelian. I'd also noted a theorem that states that nonabelian groups must have an order that is at least 5. There are two groups: Z4 and Z2 * Z2.

For order 5, the only one is Z5.

For order 6, we find prime factors 2 and 3. There is only one 3-group and either one or three 2-groups. If only one 2-group, that leaves two elements left over, and they must have the remaining order: 6. Thus, I find Z6. If three 2-groups, then that gives us Dih(3).

-

A note about dihedral and related groups.

Consider a group with order 2n and a subgroup that is Z: elements e, a, a², ..., a^n-1. The remaining elements are b, a.b, a².b, ..., a^n-1.b. Or else b, b.a, b.a², ..., b.a^n-1.

Let us see how they are related. First, b² = a^s, since b appears at most as power 1 in the elements. Also, a.b = b.a^r. Repeating, a.b.b = b.b.a^{r^2}, giving r^2 = 1 mod n. That means that r = 1 or n-1. The first one is abelian, the second one a dihedral-like group.

s = 0: Dih -- the dihedral group proper, the group of 2D rotations and reflections
s = n/2: Dic(n/2) -- the dicyclic group, a sort of modified dihedral group

For Dih(3), the elements are e, a, a², b, a.b, a².b, with a³ = e, b² = e, and a.b = b.a²

-

For order 7, the only one is Z7.

For order 8, the abelian ones are Z8, Z4 * Z2, and Z2 * Z2 * Z2. The nonabelian ones must have no elements of order 8, but elements of order 4. Thus, the nonabelian ones must be dihedral or dihedral-like groups: Dih(4) and Dic(2). Their elements are:

e, a, a², a³, b, a.b, a².b, a³.b, with a⁴ = e, a.b = b.a³, and b² either e, for Dih(4) or a², for Dic(2).

For order 9, the groups are Z9 and Z3 * Z3.

For order 10, the groups are Z10 and Dih(5).

For order 11, the only one is Z11.

 

[lpetrich]

Before doing order 12, I'll return to generalized dihedral groups. They are defined by aⁿ = e, b^m = a^s, a.b = b.a^r. From group theory, we can find the possible r and s values for each n and m values.

First we take the third equation and repeat it m times with b:
a.b^m = b^m.a^{r^m}

Since b^m commutes with a, a^{r^m} = a, giving us r^m = 1 mod n.

Then we multiply the second equation by b on both sides:
b^m+1 = b.a^s = a^s.b = b.a^r*s

Thus, s*(r-1) = 0 mod n

-

r = 1 is always a solution -- the abelian case.

For m = 2, r = n-1 is always a solution. For s = 0, we get the dihedral group, Dih, and for even n, we can get s = n/2, giving the dicyclic group Dic

 

[lpetrich]

Now for order 12.

The possible abelian groups are Z4 * Z3 = Z12 and Z2 * Z2 * Z3 = Z2 * Z6.

For nonabelian groups, we use Sylow's theorems. If there is only one 3-group and one 4-group, then the remaining elements must have order 6 or maybe also 12. Thus giving subgroup Z6, and thus the two abelian groups and two nonabelian ones, Dih(6) and Dic(3). There is an interesting curiosity: Dih(6) = Z2 * Dih(3).

If there is more than one 3-group, then there must be 4 of them. This leaves room for only one 4-group, and these subgroups exhaust all the elements (1 + 4*(3-1) + (4-1)).

Likewise, if there is more than one 4-group, then there must be 3 of them. These leaves room for only one 3-group, and with all the 4-groups' non-identity parts distinct, these subgroups also exhaust all the elements (1 + (3-1) + 3*(4-1)). If the 4-groups are versions of Z4, and if they share the order-2 element (generator^2), then there must be some elements with order 6 or 12, and we end up with the group containing Z6.

So which of these possibilities give groups? The case of 4 3-groups and 1 4-group includes A4, but I don't know about the case of 1 3-group and 3 4-groups. I'll have to work it out for both cases.

-

For the case of 3 4-groups and 1 3-group, let the first 4-group's nonidentity elements be a1, a2, a3, and the 3-group's ones are b and b². The 3 4-groups are thus (a1, a2, a3), (b.a1, b.a2, b.a3), and (b².a1, b².a2, b².a3) or mixtures of the second and third ones. For the first 4-group, one can always choose the elements so that a3 = a1.a2, whether a2 = a1² (Z4) or not (Z2*Z2).

I will consider conjugation of the 3-group by the first 4-group's elements: a1.b.a1^-1 = b or b² and likewise for a2 and a3. Or, more generally, a(k).b.a(k)^-1 = b^p(k). From a3 = a1.a2, p(3) = p(1) + p(2) mod 2. So either all three are zero or only one is zero. Meaning that b commutes with at least one of a1, a2, a3.

But b and b² can only make the 4-groups conjugate by not commuting with any of their elements, so that is a contradiction. Thus, an order-12 group cannot have three 4-groups and one 3-group.

-

This leaves the remaining case, one 4-group and three 3-groups. The nonidentity elements of the 4-group are a1, a2, a3, with a3 = a1.a2, and one of the 3-groups has nonidentity elements b and b². Let us try conjugating the 4-group by b. It should be the 4-group again, so let us see what one can come up with.

b.(a1,a2,a3).b^-1 = (a1,a2,a3)
The group is abelian.

b.(a1,a2,a3).b^-1 = (a1,a3,a2)
Repeat it:
b².(a1,a2,a3).b^-2 = (a1,a2,a3)
Thus, the group is abelian, a contradiction. This result also holds true for conjugates (a3,a2,a1) and (a2,a1,a3).

But if
b.(a1,a2,a3).b^-1 = (a2,a3,a1)
Then
b².(a1,a2,a3).b^-2 = (a3,a1,a2)
and
b³.(a1,a2,a3).b^-3 = (a1,a2,a3)
Thus this is a possible conjugacy. The other one is (a3,a1,a2), though that can be derived by taking b -> b² for the first one.

Let us consider the case of the 4-group being Z4, and let us make a1 one of its generators. Then, a2 = a1² and a3 = a1³. Thus, b.a1.b^-1 = a1[/sup]2[/sup]. Repeating this operation gives

b³.a1.b^-3 = a1⁸
or a1 = e
Which cannot happen. Thus, the 4-group must be Z2 * Z2.
b.a1.b^-1 = a2
b.a2.b^-1 = a1.a2
b.(a1.a2).b^-1 = a2.(a1.a2) = a1
So Z2 * Z2 is consistent.

I thus have shown that the 1 of 4, 4 of 3 order-12 group has a normal subgroup that is Z2*Z2.

Let us see if we can go further. With the conjugacy by b on the 4-group, let us try conjugacy of b with the a's:
a1.b.a1 = b.a3.a1 = b.a2 = a3.b
a2.b.a2 = b.a1.a2 = b.a3 = a1.b
a3.b.a3 = b.a2.a3 = b.a1 = a2.b
This uniquely determines the remaining elements of the group, so we get Alt(4).

Thus, the order-12 groups are
Z4 * Z3 = Z12
Z2 * Z2 * Z3 = Z2 * Z6
Dih(6) = Z2 * Dih(3)
Dic(3)
Alt(4)

 

[lpetrich]

The next ones are:

13: Z13

14: Z14 = Z2 * Z7, Dih(7)

15: Z15 = Z3 * Z5 (5 does not have the form 3k+1 for integer k)

16: the abelian ones alone are Z16, Z2 * Z8, Z4 * Z4, Z2 * Z2 * Z4, Z2 * Z2 * Z2 * Z2 -- 5 groups. I'll leave off, since finding all the nonabelian ones will be difficult.

 

[lpetrich]

One can find more of these groups at Wikipedia.  List of small groups has them up to order 31. It also includes how many of them there are up to order 143:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2, 2, 1, 231, 1, 5, 2, 16, 1, 4, 1, 14, 2, 2, 1, 45, 1, 6, 2, 43, 1, 6, 1, 5, 4, 2, 1, 47, 2, 2, 1, 4, 5, 16, 1, 2328, 2, 4, 1, 10, 1, 2, 5, 15, 1, 4, 1, 11, 1, 2, 1

The big ones are power-of-2 orders: 1 to 128.

The On-Line Encyclopedia of Integer Sequences® (OEIS®) is a very nice mathematical resource. You can put in a sequence of numbers and it will show which one(s) match. You can also search by keyword. It includes

A000001 - OEIS - Number of groups of order n

and it includes a list for up to 2047. The powers of 2 in it, from 1 to 1024, are:
1, 1, 2, 5, 14, 51, 267, 2328, 56092, 10494213, 49487365422


Not surprisingly, there are many fewer abelian groups. Up to order 143:
1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 7, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 5, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 11, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 5, 5, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 7, 1, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1, 6, 1, 1, 1, 5, 1, 1, 1, 2, 2, 1, 1, 3, 2, 1, 1, 2, 3, 2, 1, 15, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 1, 2, 1, 1, 1

A000688 - OEIS - Number of Abelian groups of order n; number of factorizations of n into prime powers.

Since all the finite abelian groups are known, that enables us to count them. For some order n, factor it:
n = (p1)^m1 * (p2)^m2 * ...

For each factor p^m, find all the groups. Then to get all the groups for order n, find the product of one group each for each prime factor. Here is a simple example:

Order 12 factorizes into 2² * 3¹. The groups for each factor are {Z4, Z2*Z2} and {Z3}. So one gets Z4*Z3 and Z2*Z2*Z3.

More generally, the groups for order p^m have the form product of Z(p^mk) where the mk's add up to m. The mk's are an integer partition of m, and the integer partitions are
1: 1
2: 2 11
3: 3 21 111
4: 4 31 22 211 1111
5: 5 41 32 311 221 2111 11111
...

The number of them is
A000041 - OEIS - a is the number of partitions of n (the partition numbers).

So the number of abelian groups with order n can be found with:

• Factor n
• For each prime power in n, find its number of partitions.
• Multiply them.

 

[lpetrich]

There is a complication. Many groups are product groups, what I will call reducible. The irreducible ones may nevertheless have nontrivial normal subgroups and quotient groups; I will not consider that sort of reducibility here.

So to find out how many groups are irreducible, one has to subtract out the reducible ones. I have found a way to do that by using only counts of groups. It involves finding all the factorizations of each group order:
1: 1
2: 2
3: 3
4: 4 . 2,2
5: 5
6: 6 . 3,2
I'll omit the next prime orders
8: 8 . 4,2 . 2,2,2
9: 9 . 3,3
10: 10 . 5,2
12: 12 . 6,2 . 4,3 . 3,2,2
14: 14 . 7,2
15: 15 . 5,3
16: 16 . 8,2 . 4,4 . 4,2,2 . 2,2,2,2
18: 18 . 9,2 . 6,3 . 3,3,2
20: 20 . 10,2 . 5,4 . 5,2,2
...

A001055 - OEIS - The multiplicative partition function: number of ways of factoring n with all factors greater than 1 (a(1)=1 by convention).

So for 4-groups, one has to subtract out all the products of 2-groups. Although there is only one 2-group, I will do the case of more than one for the sake of generality. For 2-groups G1 and G2, one gets G1*G1, G1*G2, and G2*G2. The case G2*G1 is equivalent to G1*G2. One might naively do

N4(all) = N4 + N2²

but that is for counting G2*G1 separately. So one has to do

N4(all) = N4 + N2*(N2+1)/2

and likewise for more repeats: N*(N+1)*(N+2)/3!, N*(N+1)*(N+2)*(N+3)/4!, etc.

Once that is done, the irreducible abelian groups are the cyclic groups with prime-power order, as one would0 expect.

That leaves plenty of irreducible nonabelian groups, and here is how many there are, a list of {order,count}:
{{6,1},{8,2},{10,1},{12,2},{14,1},{16,7},{18,2},{20,2},{21,1},{22,1},{24,6},{26,1},{27,2},{28,1},{30,1},{32,33},{34,1},{36,4},{38,1},{39,1},{40,5},{42,2},{44,1},{46,1},{48,23},{50,2},{52,2},{54,6},{55,1},{56,5},{57,1},{58,1},{60,3},{62,1},{63,1},{64,200},{66,1},{68,2},{70,1},{72,19},{74,1},{75,1},{76,1},{78,2},{80,24},{81,8},{82,1},{84,3},{86,1},{88,4},{90,2},{92,1},{93,1},{94,1},{96,121},{98,2},{100,6},{102,1},{104,5},{106,1},{108,14},{110,2},{111,1},{112,17},{114,2},{116,2},{117,1},{118,1},{120,13},{122,1},{124,1},{125,2},{126,4},{128,1999},{129,1},{130,1},{132,1},{134,1},{136,6},{138,1},{140,2},{142,1}}

Removing the dihedral and dicyclic groups gives
{{12,1},{16,5},{20,1},{21,1},{24,4},{27,2},{32,31},{36,3},{39,1},{40,3},{48,21},{52,1},{54,4},{55,1},{56,3},{57,1},{60,2},{63,1},{64,198},{68,1},{72,17},{75,1},{80,22},{81,8},{84,2},{88,2},{93,1},{96,119},{100,5},{104,3},{108,13},{111,1},{112,15},{116,1},{117,1},{120,11},{125,2},{126,2},{128,1997},{129,1},{136,4},{140,1}}

Removing the two-odd-prime nonabelian groups gives
{{12,1},{16,5},{20,1},{24,4},{27,2},{32,31},{36,3},{40,3},{48,21},{52,1},{54,4},{56,3},{60,2},{63,1},{64,198},{68,1},{72,17},{75,1},{80,22},{81,8},{84,2},{88,2},{96,119},{100,5},{104,3},{108,13},{112,15},{116,1},{117,1},{120,11},{125,2},{126,2},{128,1997},{136,4},{140,1}}

Dihedral, dicyclic, and two-odd-prime nonabelian groups are subcategories of generalized dihedral groups, and I'd have to find some way of counting them to do more.

 

[lpetrich]

Back to the generalized dihedral groups. Their conjugacy class content is:
{a^{u*r^k} for k = 0 to (m-1)} for u = 0 to n-1
{b^t*a^{u+k*(r^t-1)} for k = 0 to n-1} for u = 0 to n-1 and t = 1 to m-1

Now for that oddity of dihedral groups, Dih(2(2n+1)) = Dih(2n+1) * Z2

For Dih(2n+1), the class content is {e}, {a^k,a^2n+1-k} for k = 1 to n, and {b*a^k for k = 0 to 2n-1}

For Dih(2n), it is {e}, {aⁿ}, {a^k,a^2n-k} for k = 1 to n-1, {b*a^2k for k = 0 to n-1}, and {b*a^2k+1 for k = 0 to n-1}

Take n -> 2n+1. Then,
Dih(2(2n+1)) = {e, a²ⁿ⁺¹} * ({e}, {a^2k,a^4n+2-2k} for k = 1 to n, {b*a^2k for k = 0 to n-1})
That is, Z2 with a²ⁿ⁺¹, and Dih(2n+1) with a² and b.

This decomposition does not work for Dih(4n), and for Dic, b² = aⁿ, and not e, so Dic does not decompose.

-

The lowest orders are degenerate cases: Dih(1) = Z2, Dih(2) = Z2 * Z2, and Dic(1) = Z4

 

[lpetrich]

The commutator of the generalized dihedral group is a cyclic group generated by a^r-1. For r = 1 (abelian), it is the identity group, while for r != 1, it is Z(n/gcd(n,r-1)).

Its quotient group thus has order m*gcd(n,r-1).

The group's center is all elements a^k.b^j where k*(r-1) = 0 mod n and r^j = 1 mod n.

I've written a Mathematica program that gives the generalized dihedral groups for each n and m, giving the r and s parameters and the class content, but beyond that, I'm stumped.

 

[lpetrich]

My effort to count groups has this endpoint:  List of finite simple groups:

• The identity group
• Prime-order cyclic groups
• Alternating groups with parameter >= 5
• 16 infinite families of groups of Lie type -- matrix groups where the matrix elements are elements of some finite field GF(p^n)
• 26 sporadic groups

Simple Lie algebras are much simpler: 4 infinite families and 5 exceptional ones. The proof of them is also much, much shorter than the proof for finite simple groups, some 10,000 journal pages.

The first five of the noncyclic simple groups have orders

• 60: A5 = A1(4) = A1(5)
• 168: A1(7) = A2(2)
• 360: A6 = A1(9) = B2(2)′
• 504: A1(8) = 2G2(3)′
• 660: A1(11)

Alternating: A5 = Alt(5), A6 = Alt(6)
Lie-type: A1(q) = PSL(2,q), A2(q) = PSL(3,q)

The first two of the Lie-type "A" series are solvable:
A1(2) = Sym(3) order 6
A1(3) = Alt(4) order 12

A1(2) elements: (10,01), (01,11), (11,10), (01,10), (11,01), (01,11) with matrix multiplication modulo 2
This group is Sym(3), Dih(3), GL(2,2), SL(2,2), PGL(2,2), and PSL(2,2) -- the smallest nonabelian group
Its classes are (1001) identity, (0111, 1110) order 3, (0110, 1101, 0111) order 2

For A1(3) = PSL(2,3), I'll do its supergroups separately.

The full linear group (48):
GL(2,3): 0110, 0111, 0112, 0120, 0121, 0122, 0210, 0211, 0212, 0220, 0221, 0222, 1001, 1002, 1011, 1012, 1021, 1022, 1101, 1102, 1110, 1112, 1120, 1121, 1201, 1202, 1210, 1211, 1220, 1222, 2001, 2002, 2011, 2012, 2021, 2022, 2101, 2102, 2110, 2111, 2120, 2122, 2201, 2202, 2210, 2212, 2220, 2221

Determinant 1 (24):
SL(2,3): 0120, 0121, 0122, 0210, 0211, 0212, 1001, 1011, 1021, 1101, 1112, 1120, 1201, 1210, 1222, 2002, 2012, 2022, 2102, 2111, 2120, 2202, 2210, 2221

Projected (24):
PGL(2,3): (0110,0220), (0111,0222), (0112,0221), (0120,0210), (0121,0212), (0122, 0211), (1001,2002), (1002,2001), (1011,2022), (1012,2021), (1021,2012), (1022,2011), (1101,2202), (1102,2201), (1110,2220), (1112,2221), (1120,2210), (1121,2212), (1201,2102), (1202,2101), (1210,2120), (1211,2122), (1220,2110), (1222,2111)

Projected, determinant 1 (12):
PSL(2,3): (0120,0210), (0121,0212), (0122,0211), (1001,2002), (1011,2022), (1021,2012), (1101,2202), (1112,2221) (1120,2210), (1201,2102), (1210,2120), (1222,2111)

Notice that in the projected ones, the matrices come in pairs, the second one 2 times the first one modulo 2. Each pair is treated as one element in the group. The projected ones are the quotient groups of their originals relative to (1001, 2002).

The noncyclic simple groups get very big very quickly, and I won't try to list any of their elements, even for the smallest of them, Alt(5) = PSL(2,4) = PSL(2,5) For that one, I'd have to list 60 elements. For the next one, PSL(2,7) = PSL(3,2), I'd have to list 168 elements.

So one has to use group-theory techniques to study them, and these get arcane very quickly.

 

[lpetrich]

I'll now look at a special case of these linear groups: the orthogonal groups O(n,F) and SO(n,F) for some arbitrary field F, especially a finite one: GF(pⁿ).

I'll first look at O(2,F). This group's elements are especially simple:
R(a,b) = (a, -b; b, a)
S(a,b) = (a, b; b, -a)
with a² + b² = 1
Their determinant values:
det(R) = 1
det(S) = -1
The group SO(2,F) contains only the R's.

For characteristic 2, that is, 2*(every field element) = 0, one gets a nice reduction:
R(a,b) = S(a,b) = (a, b; b, a)
This is because of a nice little result for char 2: x + x = 2x = 0, thus -x = x.

The constraint a² + b² = 1 can be expressed as a² + 2*a*b + b² = (a + b)² = 1
Meaning that a = b + 1. Thus, the group's elements are
R(a) = R(a+1,a) = (a+1, a; a, a+1)
for all elements a of F. They have this multiplication law:
R(a).R(b) = R(a+b)

Thus, O(2,2ⁿ) = SO(2,2ⁿ) = (Z2)ⁿ

Turning to other characteristic values, like 0 for rational numbers and their supersets, one cannot do these simplifications, but there is a nice result: SO(2,F) is abelian -- for all fields, and not just numbers. This is from
R(a1,a2).R(b1,b2) = R(a1*b1 - a2*b2, a1*b2 + a2*b1)

The inverse of R(a,b) is R(a,-b), and the inverse of S(a,b) is S(a,b), another result that is also true of finite fields.

I am unable to show that SO(2,F) is cyclic, but I have found it to be true for F = GF(p) for the first 100 primes p. I have found SO(2,F) to have an order divisible by 4, an order that is either p+1 or p-1.

Here are {p, generator of SO(2,p), order of SO(2,p)} for the first eight odd primes p:
{3, {0, 1}, 4}, {5, {0, 1}, 4}, {7, {2, 2}, 8}, {11, {3, 5}, 12}, {13, {2, 6}, 12}, {17, {4, 6}, 16}, {19, {3, 7}, 20}, {23, {4, 10}, 24}

I cannot find any pattern in the generators.

If SO(2,F) is cyclic, then O(2,F) is a dihedral group. If SO(2,F) is a product of cyclic groups, then O(2,F) is the product of the corresponding dihedral ones.

So for finite fields F with characteristic not 2, SO(2,F) and O(2,F) have structures similar to finite subgroups of SO(2) and O(2).

 

[lpetrich]

For O(2,F), we have this conjugacy: S.R(a,b).S^-1 = R(a,-b) = (R(a,b))^-1. It's the dihedral group's sort of conjugacy because it was done with simple arithmetic, something that every algebraic field has. More broadly, it's interesting to see what generalizes because of the field properties of the real numbers -- and what doesn't.

All the O(n,F) contain a group which is the signed permutation matrices, those matrices where every nonzero element is 1 or -1. The latter is p-1 for GF(pⁿ). SO(n,F) selects out those matrices with determinant 1. The groups O(2,3) and O(2,5) are both the 2D version of this group, a version of Dih(4).

But for characteristic 2, that group becomes the group of the plain permutation matrices.

Now for how O(n,F) and SO(n,F) are related. For characteristic 2, like for GF(2): {0,1}, O(n,F) = SO(n,F). Otherwise, they are different. For odd n, O(n,F) = SO(n,F) * Z2 (multiplication by 1 or -1), while for even n, one does not get such a decomposition. However SO(n,F) is a normal subgroup of O(n,F) for all n, with quotient group Z2 (rotations and reflections).

 

[lpetrich]

I'll now look at the 3D case. Since (reflections) = - (rotations), I can do both at the same time. I will also do it for characteristic other than 2, because char = 2 needs to be handled separately.

The solution, for scalar part qs and vector part qv of quaternion 4-vector q:
R_ij(q) = (qs² - qv.qv)*δ_ij + 2*qv_i*qv_j + 2*qs*ε_ijkqv_k

Rotation multiplication can be found using quaternion multiplication:
(q1 x q2)s = qs1*qs2 - qv1.qv2
(q1 x q2)v = qs1*qv2 + qv1*qs2 - (qv1 x qv2)

One can find a solution for the whole matrix using its antisymmetric part, vector b. With c0 = sqrt(1 - b.b) and c1 = 1/(1 + c0) we find
R_ij(q) = c0*δ_ij + c1*b_i*b_j + ε_ijkb_k

For field F, being able to take the square root imposes some constraints on the b's.

To get to the quaternion form, we need q0 = sqrt((1 + c0)/2). This imposes further constraints.


I tried calculating some of the SO(3,F) groups.
SO(3,3) ~ Sym(4) with normal subgroup Alt(4) and quotient group Z2.
SO(3,5) ~ Sym(5) with NS Alt(5) and QG Z2
SO(3,7) ~ (?) with NS PSL(3,2) and QG Z2
Something in common:
SO(3,F) ~ some group with normal subgroup PSL(2,F) and quotient group Z2.

This suggests that the SO(3,F) groups are related to the PSL(2,F) groups -- much like the relationship of the group SO(3) to the group SU(2). By using a modification of the quaternions, I have successfully demonstrated a relationship. For every element of SL(2,F), there is a corresponding element of SO(3,F).


Going from 3D to 4D, for SO(4,F), one can find the rotations from a pair of quaternions. Each pair member does
Rs_ij(q,s) = qs*δ_ij + ε_ijkqv_k
Rs_i4(q,s) = s*qv_i
Rs_4i(q,s) = - s*qv_i
Rs₄₄(q,s) = qs
where s = +-1.
R(q1,q2) = Rs(q1,1).Rs(q2,-1) = Rs(q2,-1).Rs(q1,1) (they commute)

Reflections can be composed by multiplying rotations by S0 = diag(1,1,1,-1), thus going from SO(4,F) to O(4,F).

Here also, one must use modified quaternions when going from SL(2,F) to SO(4,F). Every pair of elements of SL(2,F) thus gives an element of SO(4,F), much like a pair of SU(2) elements giving an SO(4) element.

 

[lpetrich]

I've composed some Mathematica code for testing SU(2) -> SO(3), SL(2,F) -> SO(3,F), SU(3)*SU(3) -> SO(4), SL(2,F)*SL(2,F) -> SO(4,F)

It also ought to be possible to do counterparts to Sp(4) ~ SO(5) and SU(4) ~ SO(6), but they are more complicated.

Sp(something)? Those are the "symplectic" groups. Their matrices D satisfy D.J.D^T = J, where J = (0, -I; I, 0) is antisymmetric. This makes Sp(2n) much like SO, though still somewhat different.


When I considered O(2,F) earlier, I was naive, imposing a² + b² = 1. It can also be -1, different for char != 2.

det(R(a,b)) = a² + b²
det(S(a,b)) = - (a² + b²)
Add + or - for the sign of the determinant.

For O(2,3):

R+: 1001 2002 0210 0120
R-: 1211 2122 1121 2212
S+: 1112 2221 2111 1222
S-: 1002 2001 0110 0220

R+: Z4
R+ R-: Z8
R+ S+: Dic(2)
R+ S-: Dih(4)
R+ R- S+ S-: order 16 -- generalized dihedral?

That last one is the full group O(2,3). It has normal subgroup Z4 with quotient group Z2(R and S) * Z2(+ and -).

 

[lpetrich]

So O(2,F) has the R+ group as a normal subgroup with quotient group Z2(R and S) * Z2(+ and -)
It is thus solvable.

I'll turn to characteristic 2 for O(3,F). The matrices have the form

R = (1+a12+a13+b, a12+b, a13; . a12, 1+a12+a23+b, a23+b; . a13+b, a23 ,1+a13+a23+b)

with constraint
a12*a13 + a12*a23 + a13*a23 + b*(1 + a12 + a13 + a23) + b² = 0

I am unable to proceed any further.

 

[lpetrich]

I worked out a few of SO(3,2ⁿ):

SO(3,2) ~ Sym(3) ~ PSL(2,2)
SO(3,4) ~ Alt(5) ~ PSL(2,4)

I was unable to find a formula for going from SL(2,F) to SO(3,F) in the char = 2 case. But it looks like SO(3,F) ~ PSL(2,F) for F = GF(2ⁿ). So it looks like PSO(3,F) ~ PSL(2,F) in general, much like SO(3) ~ SU(2) / Z2(I,-I).

I note that there are also projective orthogonal groups. O(n,F) has a subgroup {a*I, aⁿ = +=1} and SO(n,F) subgroup {a*I, aⁿ = 1}. The subgroups' quotient groups are PO(n,F) and PSO(n,F), respectively.

But for the char = 2 case, we have both
GL(n,F) = SL(n,F) = PGL(n,F) = PSL(n,F)
O(n,F) = SO(n,F) = PO(n,F) = PSO(n,F)

There are also projective symplectic groups PSp(n,F)

Back to SO(n,2), I find
SO(2,2) ~ Z2 ~ Sym(2) (permutation group)
SO(3,2) ~ Dih(3) ~ Sym(3) (permutation group)
SO(4,2) ~ Z2 * Sym(4) (group of permutation matrices with 1 - (those matrices))
SO(5,2) ~ an order-720 group with a simple order-360 subgroup.

 

[lpetrich]

I finally now have a clue about what a "semidirect product" is.

A "direct product" is a simple side-by-side product: elements (a,b) for a in group A and b in group B. Their product is
(a1,b1)*(a2,b2) = (a1*a2,b1*b2)

It is easy to show that this satisfies all the group properties.

For a semidirect product, one twiddles b2 with a1:
(a1,b1)*(a2,b2) = (a1*a2, b1*f(a1,b2))

where f is some function f(A element, B element) -> B element.

The group properties gives us some constraints on f.

Associativity: f(a1, b2*f(a2,b3)) = f(a1,b2) * f(a1*a2, b3)

Identity: the element (e,e), giving f(e,b) = b and f(a,e) = e

Inverse: the element (a^-1, (f(a^-1,b))^-1) where b*f(a, (f(a^-1,b))^-1) = e

Setting b2 = e in associativity gives us concatenation: f(a1, f(a2,b)) = f(a1*a2, b)

Setting a2 = e in associativity gives us distributivity: f(a, b1*b2) = f(a,b1) * f(a,b2)

From these two identities, one finds that the second inverse identity does not give us any further constraints:
b*f(a, (f(a^-1,b))^-1) = f(a, f(a^-1,b)) * f(a, (f(a^-1,b))^-1) = f(a, f(a^-1,b) * (f(a^-1,b))^-1) = f(a,e) = e

 

[lpetrich]

A simple sort of indirect product is in the dihedral group. With elements b^k*a^m, where a² = bⁿ = e and a*b = b^n-1*a, one gets elements (a<sup>m, bk), where f(e,bk) = e and f[a,bk) = bn-k.

This is the symmetry group of a regular n-gon. A rotation advances the vertices while wrapping them around, and a reflection reverses their order, and may also advance them. A small backward rotation is an almost-complete forward rotation.

Triangle: rotations 123 231 312, reflections 321 213 132
Square: rotations 1234 2341 3412 4123, reflections 4321 3214 2143 1432
Etc.


Another one is the n-D orthoplex and hypercube symmetry group. A hypercube generalizes the square and the cube to more space dimensions, and an orthoplex does likewise with the square and the octahedron. Hypercube vertices: (+-1, +-1, ..., +-1), orthoplex vertices: (all 0 but one, which is +-1).

For coordinate-aligned ones, their symmetry group is the permutation matrices for n symbols, where the 1's in it become +-1's. The permutations rearrange the axes, while the +-1's flip along the axes. This group has order 2n*n!. It is the semidirect product of Sym for the permutation matrices, and (Z2)n for the +-1's. Its elements are (permutation, +-1's vector), where f(permutation, vector) is (apply permutation to vector).


More generally, the functions f do automorphisms on B, though different a's may do the same automorphism.


So for a direct product, one can define homomorphisms F((a,b)) = a and F((a,b)) = b, or F(A*B) = A and F(A*B) = B, while for semidirect products, one only has F((a,b) = a. But one also has F((a,b)) = automorphism f of a on B.</sup>

 

[lpetrich]

The group of automorphisms of a group G is called Aut(G).

It has a subgroup, the "inner automorphisms", Inn(G), which are automorphisms generated by conjugation: for some x, a -> x*a*x^-1. Inn(G) is equal to the quotient group G / Z(G), where Z(G) is the "center" of G, all the elements of G that commute with every other element of G. Needless to say, Z(G) is a subgroup of G.

The outer automorphisms are all the others, and there is an outer-automorphism group that is given by Out(G) = Aut(G) / Inn(G).


Let's look at some examples. For abelian groups, the only inner automorphism is the identity one, making every nontrivial one an outer one.

Cyclic ones have a rather simple sort of automorphism: a -> a^s for any s that is relatively prime to the order of the group.

I'll derive the possible automorphisms of the dihedral groups Dih for n >= 3. They are generated by a and b such that a² = bⁿ = e and a*b = b^n-1*a.

Since b has order n, it must be mapped onto an element with order equal to n, meaning that it cannot be mapped onto a*b^m for any m -- those elements have order 2. It must be mapped onto b^s for some s relatively prime to n. This takes care of all the powers of b in the group, meaning that a must be mapped onto some a*b^t for some t. The final task is the non-commutation identity:
a*b -> a*b^s+t
b^n-1*a -> b^s(n-1)*a*b^t = a*b^s+t
Meaning that every relatively-prime s and every t gives an automorphism. Its order is thus n*phi, where phi is Euler's totient function, the number of positive integers <= n that are relatively prime to n.

Let us now consider the inner automorphisms. For odd n, Dih's center is the identity, and every element generates an automorphism. For even n, Dih has center {e,b^n/2}, so every automorphism is generated by both some element and that element multiplied by b^n/2.

For Dih(3), |automorphisms| = 6, and |inner ones| = 6 also, so every one is an inner one. But for Dih(4), |automorphisms| = 8 and |inner ones| = 4. with a Z2 quotient group.

For Dih(3),
a -> a, b -> b: conjugation by e
a -> a, b -> b²: conjugation by a
a -> a*b, b -> b: conjugation by b
a -> a*b, b -> b²: conjugation by a*b²
a -> a*b², b -> b: conjugation by b²
a -> a*b², b -> b²: conjugation by a*b
In summary,
a -> a, a*b, a*b², and b -> b, b²
covering all six automorphisms.

For Dih(4),
a -> a, b -> b: conjugation by e
a -> a, b -> b³: conjugation by a
a -> a*b², b -> b: conjugation by b
a -> a*b², b -> b³: conjugation by a*b
a -> a, b -> b: conjugation by b²
a -> a, b -> b³: conjugation by a*b²
a -> a*b², b -> b: conjugation by b³
a -> a*b², b -> b³: conjugation by a*b³
In summary,
a -> a, a*b², and b -> b, b²
With these four, there are four additional ones;
a -> a*b, a*b³ and b -> b, b³

 

[lpetrich]

I'd mentioned regular polygons and polyhedra and their generalizations in more space dimensions, and I will go into a bit more detail about that.

We start with the polygons. An n-gon has n vertices and n edges, and a regular one has symmetry group Dih, with order 2n:

What	Vertices	Edges	Group
n-gon	n	n	2n

Vertices have zero dimensions, and edges one dimension. Going to polyhedra, we add faces, two-dimensional objects:

What	Vertices	Edges	Faces	Group
Tetrahedron	4	6	4	24
Cube (hexahedron)	8	12	6	48
Octahedron	6	12	8	48
Dodecahedron	20	30	12	120
Icosahedron	12	30	20	120

The names of these polyhedra are essentially 4-face, 6-face, 8-face, 12-face, and 20-face.

Going to four dimensions, we get polychora (singular polychoron), and we add cells, three-dimensional objects:

What	Vertices	Edges	Faces	Cells	Group
Pentachoron	5	10	10	5	120
Tesseract (octachoron)	16	32	24	8	384
Icositetrachoron	24	96	96	24	1152
Hexadecachoron	8	24	32	16	384
Hecatonicosachroron	600	1200	720	120	14400
Hexacosichoron	120	720	1200	600	14400

The more usual names for these polychora are 5-cell, 8-cell, 24-cell, 16-cell, 120-cell, and 600-cell.

The most general name for these shapes is polytope, and in more than four dimensions, there are only three types: the simplex, the hypercube, and the orthoplex or cross-polytope.

I'll use B(n,m) for n!/(m!*(n-m)!)) -- the binomial-theorem coefficients. I also note that a zero-dimensional regular polytope is a point and one-dimensional one a line segment.

The n-simplex has (n+1) equally distant vertices, thus generalizing the triangle, the tetrahedron, and the 5-cell. Its dimension-k subpolytopes are B(n+1,k) k-simplexes. It is self-dual. Its symmetry group is the group of all permutations of the vertices, Sym(n+1). Its order is (n+1)!

The n-hypercube has 2ⁿ vertices at positions +-1 in all coordinates, thus generalizing the square, the cube, and the tesseract. Its dimension-k subpolytopes are 2^n-k*B(n,k) k-hypercubes.

The n-orthoplex or n-cross-polytope has 2n vertices with coordinate values all zero except for one, with value +-1. Its dimension-k subpolytopes are 2^k+1*B(n,k+1) k-orthoplexes.

The n-hypercube and n-orthoplex are duals of each other, and they share a symmetry group that I'd mentioned earlier. Its elements are permutation matrices with the 1's turned into +-1's, making it the semidirect product of n-permutations and n-vectors of +-1. Its order is 2ⁿ*n!


There are five sporadic regular polytopes, five that occur outside of the four infinite families of them. The three-dimensional ones are the dodecahedron and the icosahedron, both duals of each other, and both having the icosahedral group as their symmetry group. The four-dimensional ones are the 24-cell, the 120-cell, and the 600-cell. The 24-cell's vertices have as coordinates two zeros and two +-1's. It is self-dual, and its symmetry group is the 16-cell's one with nonsingular matrices with elements +- 1/2. That group has order 1152. The 120-cell and the 600-cell are duals of each other, and they share a 14400-element symmetry group that is approximately a product of two icosahedral groups.