The Math Thread

[Bomb#20]

I tried doing this calculation for fullerene with Ih -- icosahedral rotations and reflections. I won't go into detail, but I got the same numbers of degeneracies as with plain I -- icosahedral rotations.

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There is a connection between the tetrahedral, octahedral, and icosahedral groups and the five Platonic solids.

The tetrahedral group is the symmetry group of the regular tetrahedron.

...

You appear to be writing down everything you know about group theory in one massive drawn-out core-dump. Did somebody challenge you to a duel?

 

[lpetrich]

You appear to be writing down everything you know about group theory in one massive drawn-out core-dump. Did somebody challenge you to a duel?

I get it.

 

[beero1000]

I tried doing this calculation for fullerene with Ih -- icosahedral rotations and reflections. I won't go into detail, but I got the same numbers of degeneracies as with plain I -- icosahedral rotations.

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There is a connection between the tetrahedral, octahedral, and icosahedral groups and the five Platonic solids.

The tetrahedral group is the symmetry group of the regular tetrahedron.

...

You appear to be writing down everything you know about group theory in one massive drawn-out core-dump. Did somebody challenge you to a duel?

:laughing-smiley-014

I'm starting to prep for my algebra course next semester and will probably post my lecture notes when they're ready. No spoilers!

 

[beero1000]

FYI, this linear algebra video youtube series is currently ongoing. So far, a bunch of the techniques and visualizations he uses overlap with my own preferred explanations. I like it.

+1 for matrix multiplication as composition.

 

[lpetrich]

I got killed in a duel. Only kidding. I'm back.

It's good to have a visually intuitive way of justifying matrix multiplication. I also like its take on determinants. I can see that that author hasn't gotten to eigenvalues and eigenvectors yet.

A good application of linear algebra is coordinate transformations. From an object's internal coordinates to the scene's overall coordinates to the view coordinates. One could simplify it when starting out by only doing 2D and by making scene coordinates = view coordinates.

For eigenvalues and eigenvectors, one can do principal components analysis, explained by fitting the data points to an ellipsoid and then finding that ellipsoid's axes. Also, one can do coupled oscillations, like two pendulums connected by a spring.

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Now some more group theory. A book on group theory has this exercise. For groups with order p²*q for primes p and q, which of its Sylow p- and q-subgroups will be normal?

I'll first do the p*q case where q > p without loss of generality. The numbers of Sylow p- and q-subgroups must satisfy
kp*p+1 = 1 or q and kq*q+1 = 1 or p
Since q > p by hypothesis, kq must equal 1, making the Sylow q-subgroup normal. The Sylow p-subgroup need not be normal if kp*p+1 = q, a condition on q.

For the p²*q case,
kp*p+1 = 1 or q and kq*q+1 = 1, p, or p²

If p > q, then the Sylow p-subgroup must be normal.
If p < q, then let us consider kq*q+1 = p²
If p is odd, then p = 2*r+1. Then kq*q = 4*r*(r+1). This means that kq must be divisible by 8: kq = 8*kq'. This gives us kq'*q = r(r+1)/2, where q must divide either r or r+1. This is contrary to the premise that p < q, so in this case, the Sylow q-subgroup must be normal. It may also be normal for p = 2.
But if p = 2, then the Sylow q-subgroup need not be normal, and if it isn't, there is a conjugate set of 4 of them. This means that q = 3 and that the order is 12. All the non-identity elements of the Sylow 3-subgroups make up 8 elements, leaving 4 for the Sylow 2-subgroup, making it normal.

So if p < q, then either the Sylow q-subgroup is normal or else the group's order is 12 with a non-normal Sylow 3-subgroup.

 

[George S]

FYI, this linear algebra video youtube series is currently ongoing. So far, a bunch of the techniques and visualizations he uses overlap with my own preferred explanations. I like it.

+1 for matrix multiplication as composition.

And here I thought I had learned linear algebra 50 years ago.

 

[lpetrich]

Some more group theory. Consider two normal subgroups A and B of a group and calculate those group's commutator. All values of a*b*a^-1*b^-1 with a in A and b in B. With associativity,
a*b*a^-1*b^-1 = (a*b*a^-1)*b^-1 = b'*b^-1 is in B
a*b*a^-1*b^-1 = a*(b*a^-1*b^-1) = a*a'^-1 is in A
Thus the commutator of A and B is in the intersection of those two subgroups.

If their only shared element is the identity, then we have
a*b*a^-1*b^-1 = e
a*b*a^-1 = b
a*b = b*a

The two subgroups commute with each other, even though they need not be commutative internally.

If all a group's Sylow subgroups are normal, then that group is a direct product of those subgroups.

Now for groups with order p²*q² where q > p.

Is the Sylow q-subgroup normal? If not, then k*q+1 = p² since q > p. This means that p² = 1 mod q, and that p = +-1 mod q. Since p is less than q, p = q - 1. There is only one combination of p and q that brings that about: p = 2 and q = 3. So we have order = 36 or the Sylow q-subgroup is normal. For order 36, at least one of the Sylow 2-subgroup or the Sylow 3-subgroup is normal (Consequences of the Sylow Theorems - sylowapp.pdf).


It can be shown that Dih(2*n) = Z2 * Dih if n is odd. More generally, Dih(2*n) -(Dih)-> Z2.

 

[lpetrich]

Fun fact: if an abstract-algebra ring's additive and multiplicative identities are equal, those equal identities are its only element.

Proof. Call that element y and suppose that that ring has element a. Then,
a*a = a*(a+y) = a*a + a*y = a*a + a
with the result that a = y.

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Now for how to get a group out of a commutative monoid (associativity, identity). There is a way, the Grothendieck construction, a way that generalizes subtraction. In fact, one can use it to get integers from natural numbers.

Call the operation + and the identity 0. Construct pairs of elements (a,b) and define operation
(a,b) + (c,d) = (a+c,b+d)
and equality
(a,b) = (c,d) if there is some x that makes c = a+x and d = b+x or else a = c+x and b = d+x
Also identify (a,0) with a.

Associativity: yes
Identity: (0,0)
Inverse of (a,b): (b,a)

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One can do something similar for multiplication. From an integral domain, a commutative ring where nonzero elements always have nonzero products, one can define a "field of fractions" with pairs of elements, a generalization of rational numbers:

(a,b) + (c,d) = (a*d+b*c,b*d)
(a,b) * (c,d) = (a*c,b*d)
(a,b) = (c,d) if a*d = b*c
a identified with (a,1)

Associativity: yes
Additive identity: (0,1)
Multiplicative identity: (1,1)
Additive inverse: (-a,b)
Multiplicative inverse: (b,a)
Distributivity: yes
No division by zero: (a,0) not allowed

One can use it to get the rational numbers from the integers.

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One can also define a generalization of algebraic numbers, by extending a ring with the solution of some polynomial equation with its coefficients in that ring. Thus, one can extend the ring of integers with a solution of x²+1 = 0 to get the ring of complex integers.

 

[lpetrich]

In the ring of complex integers, factoring is not unique.
2 = (1+i)*(1-i) -- 2 is no longer a prime
10 = 2*5 = (3+i)*(3-i) -- (1+3*i)*(1-3*i) is a variation of (3+i)*(3-i)

Of positive integers with forms 4*k, 4*k+1, 4*k+2, and 4*k+3, which ones cannot have complex-integer factors? Of those forms that can have non-real complex-integer factors, which numbers with those forms are the smallest counterexamples in each one?

 

[lpetrich]

This problem turns out to be overly simple. Consider:
4k -> 4 = 2² = 2*(1+i)*(1-i) = (1+i)²*(1-i)²
4k+1 -> 25 = 5² = 5*(2+i)*(2-i) = (2+i)²*(2-i)²
4k+2 -> 6 = 2*3 = 3*(1+i)*(1-i)
4k+3 -> 15 = 3*5 = 3*(2+i)*(2-i)

So I'll pose a more restricted problem:
Of the various classes, 4k, 4k+1, 4k+2, 4k+3, which have no members that can be expressed in the form (a+b*i)*(a-b*i)? Of those that have members that can, which member is the smallest counterexample?

 

[lpetrich]

I'll solve that problem.

(a+i*b)*(a-i*b) = a² + b²

For a = 2*ka and b = 2*kb, a² + b² = 4*(ka² + kb²)
For a = 2*ka and b = 2*kb+1, a² + b² = 4*(ka² + kb² + kb) + 1
For a = 2*ka+1 and b = 2*kb+1, a² + b² = 4*(ka² + kb² + ka + kb) + 2

So classes 4*k, 4*k+1, and 4*k+2 can potentially be factored in this fashion. Class 4*k+3 cannot be.

The first ones that cannot be:
4*k: 12
4*k+1: 9
4*k+2: 6

The 4*k case reduces to the other cases in this fashion: 4*k = a² + b² -> k = (a/2)² + (b/2)² repeated until one gets a number that cannot be evenly divided by 4.

There's an interesting theorem that states that every prime with form 4*k+1 can be expressed as the sum of two squares. That is, it can be factored in this fashion.  Fermat's theorem on sums of two squares,  Proofs of Fermat's theorem on sums of two squares,  Sum of two squares theorem, Fermat's Two Squares Theorem - ProofWiki

Furthermore, this decomposition for such primes is unique, though it is not unique for some composite numbers.

 

[lpetrich]

4*k+1: 9 -- that's only correct if one excludes 0, since 9 = 3² + 0²
Allowing 0, the first non-decomposable one is
4*k+1: 21

In general, a positive integer cannot be decomposed into the sum of two squares if it has at least one odd power of its prime factors that cannot be decomposed in that fashion.

This property of multiple factors is not exclusive to the complex integers. It is also the case for the ring Z(sqrt(2)), the integers extended with the square root of 2: a + b*sqrt(2).

1 = (sqrt(2)+1)*(sqrt(2)-1) = (3+2*sqrt(2))*(3-2*sqrt(2)) = ...

The second factorization is the square of the first one.

Likewise, for Z(sqrt(3)),
1 = (2+sqrt(3))*(2-sqrt(3)) = ...

and for Z(sqrt(5)),
1 = (sqrt(5)+2)*(sqrt(5)-2) = ...

and for Z(sqrt(6)),
1 = (5+2*sqrt(6))*(5-2*sqrt(6)) = ...

Now for extending Z with a rational number: Z(1/2), elements a + b/2.
1 = 2*(1/2) = ...

These extensions of the integers are all integral domains (commutative, no zero divisors), but not unique factorization domains.

 

[lpetrich]

Now for automorphisms of algebraic fields -- this figures in the theory of why only some polynomial equations can be solved with radicals (nth roots).

Let's consider the finite fields first. The fields GF(p) have only the identity automorphism, since all their nonzero elements can be generated from 1.

But fields GF(p²) and higher powers can have nontrivial automorphisms. Consider GF(4). Its elements are 0, 1, x, x+1, where x satisfies x²+x+1 = 0

Consider x -> x+1. x²+x+1 -> (x+1)² + (x+1) + 1 = x²+2x+1+x+1+1 = x²+x+1.

It is thus an automorphism of GF(4). It and the identity automorphism form a group with abstract group Z2.

In general, a finite field has a "Frobenius automorphism" x -> x^p. It generates a cyclic group of automorphisms with order n, where the field is GF(pⁿ).

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A "splitting field" is the smallest field that contains a polynomial's roots, and that is also an extension of the field that the polynomial's coefficients are defined over.

Finite field GF(pⁿ) is thus the splitting field of x^{p^n} - x over GF(p)

For polynomial x² - 3*x + 2 over the rational numbers Q, the splitting field is also Q, since roots 1 and 2 are in Q.

But for polynomial x² - 2 over the rational numbers Q, the splitting field is Q(sqrt(2)), since sqrt(2) is not a rational number.

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Now for the automorphisms of the integers, the rational numbers, the algebraic numbers, the real numbers, and the complex numbers.

The ring of integers is easy. An automorphism must preserve the ring properties: f(a+b) = f(a)+f(b), f(a*b) = f(a)*f(b), f(0) = 0, f(-a) = -f(a), f(1) = 1.

Thus, f(-1) = -1, and if f(a) = a, then f(a+1) = a+1 and f(a-1) = a-1, and by mathematical induction, f(a) = a for all a. Thus, the ring of integers has only the trivial automorphism, the identity one.

For the field of rational numbers, we consider (a/b) * b = a, where a and b are rational numbers identified with integers.
f(a/b) * f(b) = f(a)
gives
f(a/b) * b = a
giving
f(a/b) = a/b

Thus, rational numbers have only the trivial automorphism. Real numbers also have only the trivial automorphism, since each real number can be expressed as a Cauchy sequence of rational numbers. That shows us that real algebraic numbers also have only the trivial automorphism.

But if we extend the rational numbers by some polynomal's roots, instead of with all polynomials' roots (that makes the algebraic numbers), we get something interesting.

Let's consider the complex rational numbers: Q(i). Take their complex conjugate, i -> -i. It satisfies the same addition and multiplication operations as the original. Thus, complex conjugation is a nontrivial automorphism, and with the trivial one, the two automorphisms form group Z2.

That is true of complex numbers in general ("complex real numbers"?).

For the rational numbers extended with sqrt(2), Q(sqrt(2)), we get automorphism sqrt(2) -> -sqrt(2). It also forms group Z2. However, the real algebraic numbers and the real numbers do not get extended, since sqrt(2) is already in them.

 

[lpetrich]

Now for some interesting properties of extension fields.

The automorphisms of an extension field E that leave the base field F fixed form its "Galois group", after that ill-fated dueler: Gal(E/F).

There is a valuable theorem: if one has a field D that is an extension of field E, and that in turn an extension of field F, then their Galois groups are related as follows:

Gal(E/F) is a normal subgroup of Gal(D/F), with quotient group Gal(D/E):
Gal(D/F) -( Gal(D/E) )-> Gal(E/F)

Or
Gal(D/F) -( Gal(D/E) )-> Gal(E/F) -( Gal(E/F) )-> (identity group: {e})

If E is a splitting field of polynomial P in F, D a splitting field for Q in E, and D a splitting field for R in F, then polynomial R is a composite of polynomials P and Q.

So if one finds that polynomial's Galois group has a nontrivial normal subgroup, then one knows that that polynomial is a composition of two other ones. In fact, if that group has a series of such subgroups, each one a normal subgroup of the previous one, then that polynomial is a composition of several ones.

This will be useful for determining whether or not a polynomial can be solved with radicals (nth roots), or even a special case of them, square roots.

 

[lpetrich]

So let us find the Galois groups of polynomial xⁿ - a over some field F.

The solutions are x_k = a^1/n * w^k where w is a nth root of unity.

If a^1/n is in F, then we need only find the Galois group of the set of the nth roots of unity. This is the group of automorphisms w -> w^b where b is relatively prime to n. The b's are all positive integers relatively prime to n, and under multiplication modulo n, they form a group. For n a prime, they form group Z(n-1), otherwise, they form smaller groups, sometimes products of cyclic groups.

The number of possible b's is given by the "Euler totient function", phi, and that is equal to n * product over prime factors p of (1 - 1/p).


If a^1/n is not in F, then the Galois group is more complicated. The roots are related by x<sub>k[/sup] = x0 * (x1/x0)^k

It has automorphisms xk = xa * (xa+b/xa)^k for some a and b. This is equivalent to
(new k) = a + b*k mod n

As with the nth roots of unity, b must be nonzero and relatively prime to n. However, a can be any of 0 ... n-1.

Combining transformations (a,b) and (a',b') gives (a+b*a',b*b').

That set of transformations is, of course, a group, and it can be shown that all elements (a,1) form a normal subgroup in it. That makes the b's group a quotient group of it.

For n = 2, there is only (0,1) and (1,1) -- identity and interchange, all permutations of roots

For n = 3, there is (0,1), (1,1), (2,1), (0,2), (1,2), (2,2) -- the group Sym(3) ~ Dih(3), all permutations of roots

For n = 4, there is (0,1), (1,1), (2,1), (3,1), (0,3), (1,3), (2,3), (3,3) -- the group Dih(4). It is a smaller group than Sym(4), all permutations of roots.

In general, for n > 3, this group, with order n*phi, will have fewer elements than the group of all permutations of roots, Sym, order n!.

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Let's resolve an issue with quotient groups that themselves contain quotient groups. Consider group G with normal subgroup S and quotient group Q. In turn, Q has normal subgroup QS and quotient group QQ.

The elements of G, a, can be expressed as a(b), a tagged with b, where b is in Q. The elements of S are thus a(e). Let's consider the tagging of elements b: b(c). b(e) will be in QS and c in QQ. Then G's elements have form

a(b(c)) or a(b,c)

One gets an intermediate subgroup T, one that also is normal: all a(b,e) where the b's are those tagged with e. One then finds that G/T = QQ and T/S = QS. Thus pushing Q's quotient group into the main series of quotient groups.

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Thus, if a polynomial can be solved using radicals / nth roots, then its Galois group can be decomposed into subgroups ending at the identity group with a series of quotient groups that are all cyclic groups.

The "composition series" of a group is a series of subgroups of it with each one being the maximal or largest proper subgroup of the previous one. It ends at the identity group. If the quotient groups along the way are all cyclic groups, then the group is "solvable".

So one finds that if a polynomial can be solved with radicals, then its Galois group is solvable.</sub>

 

[lpetrich]

Let us now put these results to work in finding out which polynomials can be solved with radicals.

The maximal Galois group of a polynomial with degree n is Sym, the symmetric group.

For quadratic equations, n = 2 and its maximal Galois group is Sym(2) ~ Z2
One needs a square root.

For cubic equations, n = 3 and its maximal Galois group is Sym(3) -( Z2 )-> Z3
One needs a cube root.

For quartic equations, n = 4 and its maximal Galois group is Sym(4) -( Z2 )-> Alt(4) -( Z3 )-> Z2*Z2
Or Sym(4) -( Sym(3) )-> Z2*Z2
One needs a cube root and two square roots, and that is what the general quartic formula contains.

For quintic equations, n = 5 and its maximal Galois group is Sym(5) -( Z2 )-> Alt(5)

But Alt(5) is "simple", meaning that it has no proper normal subgroups. Thus, the general quintic polynomial cannot be solved with radicals, though it's easy to find special cases that can be.

This is also true of all higher degrees of polynomial: Sym -( Z2 )-> Alt -- Alt is simple for n >= 5. A simple group G has composition series G -( G )-> {e}

 

[lpetrich]

Now for ruler-and-compass solutions. These involve arithmetic operations and square roots, and for some quantity to be found using ruler and compass, its polynomial's Galois group must contain only Z2's in its composition series. Such cases as Z4 and Z2*Z2 do so: Z4 -( Z2 )-> Z2 and Z2*Z2 -( Z2 )-> Z2.

For trisecting a general angle, one must solve a general cubic polynomial, and its composition series contains Z3 along with Z2. So it is not possible to trisect the angle with ruler and compass. The same is true of duplicating the cube, where one must find 2^1/3.

But if one uses a "neusis" or marked ruler, one can solve cubic polynomials, and if a polynomial's composition series contains nothing but Z2's and/or Z3's, then one can do a neusis construction. Thus, one can solve cubic and quartic polynomials with a neusis, and trisect an angle and duplicate the cube with it.

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Let's see what regular polygons can be constructed. The center and two neighboring vertices form an isosceles triangle that can be split into two mirror-image right triangles. The angle at the center is pi/n for a n-gon. So we must find cos(pi/n). It solves T(n,cos(pi/n)) = -1 where T is the first kind of Chebyshev polynomial.

If n has factors p1^k1*p2^k2*... then 1/n can be expressed as m1/p1^k1 + m2/p2^k2 + ...

That means that we only need to consider powers of prime factors.

For ruler-and-compass construction, powers of 2 are easy. All they need is square roots. For powers of other primes, we must consider their relatively-prime-number groups. These have order phi(p^k) = p^k-1*(p-1). So only primes with the form 2^k+1 can appear, and each one can only appear once. Going further, the only possible primes with that form are 2^{2^k}+1, the Fermat primes.

So to be constructed with ruler and compass, n = 2^m * at most one of each Fermat prime.

The only known Fermat primes are 3, 5, 17, 257, and 65537, corresponding to k = 0, 1, 2, 3, 4. It is unknown whether there are any others, though all other "Fermat numbers" that have been tested for primality have been shown to be composite.

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If one allows a neusis construction, one allows cube roots, and one can thus have primes of form 2^k1*3^k2+1. Thus, the possible values of n = 2^m1 * 3^m2 * at most one of each prime with the above form.

With primes with that form, one can construct regular polygons with n = 5, 7, 13, 17, 19, 37, 73, 97, 109, 163, 193, 257, ... sides

If one can do a quintic-solving construction, then one can use primes of form 2^k1*3^k2*5^k3+1. Primes that include 7, 11, 13, 17, 19, 31, 37, 41, 61, 73, 97, 101, 109, 151, 163, 181, 193, 241, 251, 257, ...

 

[lpetrich]

Some clarification.

For a cubic (degree-3) polynomial, the general formula for its roots involves a square root and a cube root. This parallels the composition series for its maximal Galois group:
Sym(3) -( Z2 )-> Z3 -( Z3 )-> {e}

For a quartic (degree-4) polynomial, the general formula for its roots involves a general cubic equation, and not simply cube roots. Thus,
Sym(4) -( Sym(3) )-> Z2*Z2 -( Z2 )-> Z2 -( Z2 )-> {e}

The Sym(3) is for the general cubic equation's roots.

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Primes with the form 2^k1*3^k2+1 are called Pierpont primes. More generally, those are Pierpont numbers. One can generalize them to other selections of primes.

One has to have a 2 among the primes to make primes, because that is the only even prime.

For Pierpont numbers, one gets 2 for k1 = k2 = 0, and one can get other primes with k1 > 0. If gcd(k1,k2) has odd factors > 1, then the Pierpont number is composite. So to make a prime, gcd(k1,k2) must be some power of 2, including power 0.

 

[lpetrich]

Some rotation-reflection groups, like the tetrahedral and octahedral ones, can be expressed in a rather simple form.

A permutation matrix is a matrix that is a permutation of the rows or columns or the identity matrix. Each row and column contains one 1 and all the rest 0.

For those rotoreflection groups, some or all of the 1's in some of the elements may be replaced by -1. Its elements are (diagonal matrix of 1's and -1's) . (permutation matrix)

This can be expressed more abstractly as (A,P) where A is a vector of elements of some group and P is a permutation. Their multiplication law is (A,P) * (B,Q) = (A*(P.B),P.Q). This is called a "wreath product".

A wreath product has a subgroup: all elements with the form (A,I), where I is the identity permutation. This subgroup is normal, and its quotient group is the group of the permutations in the group.

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Now for the symmetry groups of the regular polytopes, generalizations of the regular polygons and polyhedra.

2 space dimensions: a regular n-gon has symmetry group Dih, with its pure rotations being Cyc.

For more space dimensions, there is only a finite number of regular polytopes for each one, and only three of them except for 3D and 4D. Here are those three.

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The simplex. It generalizes the 2D triangle, the 3D tetrahedron and the 4D 5-cell. For n-D, it has (n+1) vertices with constant separation between them. It is self-dual, and its symmetry group is the n-D representation of Sym(n+1) (rotoreflections) and Alt(n+1) (rotations).

The hypercube. It generalizes the 2D square, the 3D cube, and the 4D 8-cell. For n-D, it has 2ⁿ vertices, at locations (+-1,+-1, ...,+-1).

The cross-polytope. It generalizes the 2D square, the 3D octahedron, and the 4D 16-cell. For n-D, it has 2n vertices, at locations (0,0,...,+-1,...,0,0)

The hypercube and the cross-polytope are duals, and they share a symmetry group: the wreath product of (n of +-1's) and all n-D permutation matrices (group Sym). It thus has 2ⁿ*n! elements. The pure rotations have all of those with determinant = +1, (+-1's multiplying to 1) . (even permutations) + (+-1's multiplying to -1) . (odd permutations)

This rotoreflection group has index-2 subgroups (pure rotations in it), (+-1's multiplying to 1), (even permutations), and an index-4 subgroup with quotient group Z2*Z2: (+-1's multiplying to 1) . (even permutations).

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In 3D, there are two additional polyhedra, duals of each other, the dodecahedron and the icosahedron. Their symmetry group is the icosahedral group, Ih (120 elements) for rotoreflections and I (60 elements) for rotations only.

In 4D, there are three additional polyhedra. One of them, the 12-cell, is self-dual and halfway between the 8-cell and the 16-cell. Its symmetry group is the 8-cell/16-cell one with elements (1/2)*(4*4 array of +-1's) added, if those elements have determinant +-1. That group has 1152 elements in total and 768 for pure rotations. This builds on the 384 elements (rotoreflection) / 192 elements (rotation) for 8-cell and 16-cell symmetry.

The other two, the 120-cell and the 600-cell, are related to the 3D dodecahedron and icosahedron. They share a symmetry group that is roughly the square of the icosahedral group, one with 14400 elements for rotoreflections and 7200 elements for pure rotations.

 

[lpetrich]

After all this stuff on finite groups, I'll turn to infinite ones.

I'll start with discrete ones, and of these ones, I'll start with the group of integers under addition: Z. It has subgroups a*Z -- a*(every integer) under addition, for integer a. Their quotient group is Z(a) -- integers modulo a under addition. In fact, I've seen Z(a) written Z/a*Z.

Z's irreps are a^k for element k and arbitrary nonzero complex a.

Z's values form a one-dimensional lattice. (Z)^d makes a d-dimensional lattice. That is important for crystal structure, because atoms in a crystal are arranged in patterns that repeat themselves over a 3D lattice.

A crystal lattice has the form x = sum over m of x(m)*k(m) where the x(m) are the unit-cell vectors and the k(m) are integers. The x(m) need not be orthogonal -- only non-collinear (nondegenerate).

Suitable translations (shifts) along a crystal's axes form a symmetry group, and one can ad rotations and reflections. The general element (D,R) (translation, rotoreflection) does: x' = R.x + D

That has the multiplication law (R1,D1)*(R2,D2) = (R1.R2, D1+R1.D2)

The elements D have the form D(lattice) + D(R), where D(R) is specific to each R value, including D(I) = 0.

Thus group contains a normal subgroup (I,D) with translations only, and a quotient group containing all the R's.

There is an interesting result that if the lattice and the rotoreflections have the same number of dimensions, there is a finite number of these groups.