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Group action - solutions.
The wrong statement was the second part of problem 4.
1. (a) Assume that prime number p divides x16 + 1. Then p is of form 32k + 1.
(b) Without using the general Dirichlet theorem or L-functions, prove that for any
k there is infinite set of primes of form kn + 1.
Solution. (a) Modulu p, the equation x16 = –1 has a solution. So, there is a
remainder x mod p, which is of order 32. By Fermat little theorem, order of each
element mod p divides p – 1, therefore p – 1 = 32k.
(b) We shall need the following notion:
Definition. Cyclotomic polynomial is.
In other words, it is a monic polynomial that has simple roots which are “roots of 1
of degree n precisely”.
A simplest way to construct it is to take xn – 1 and divide it by all common divisors
with xk – 1 for k < n. From this it can be seen that cyclotomic polynomial has
integer coefficients.
Generic remarks on cyclotomic polynomials:
a. If You write down first one-hundred-something cyclotomic polynomials, you
will get an impression that all their nonzero coefficients ±1. That impression is
wrong.
b. It can be shown that all cyclotomic polynomials are irreducible. But that’s an
exercise with 2 stars, and we won’t discuss it right now.
c. Cyclotomic polynomials have many applications, for example to geometry of
regular polygons and to Galois theory.
Back to the original problem. Assume that there is only a finite number of primes
of type kn + 1. Let P be the product of all those primes.
Consider the number  n  P  . It is not divisible by the divisors of P, since the free
coefficient of cyclotomic polynomial is 1 for n > 1 (why?). But it has a prime
divisor, p. We claim that p is of form kn + 1, which contradicts the assumption.
Indeed,  n  P   0  mod p  hence P n  1 mod p  and for no m < n it we shall get
P m  1 mod p  because  n  x  , x m  1 are coprime polynomials.
Therefore, the order of P mod p is n and by Fermat little theorem, n divides p – 1.
2. (a) Pizza consists of p sectors (p is a prime number). Each triangle can be one of
a types (onions, olives, mushrooms and so on). Compute the number of possible
nonequivalent pizzas.
(b)* Let p be a prime number. Compute the quantity of subsets of p elements of a
set {1, 2, 3, …, 2p} such that sum of subset elements is divisible by p.
2p
 p 2
ap  a
2
 a (b)  
Answers. (a)
p
p
Remark. From these one can get nice conclusions about divisibility.
From (a) we get Fermat little theorem: a p  a  mod p 
 np   n 
Obvious generalization of (b) is       mod p 
 kp   k 
Solution. (a) If pizza wouldn’t rotate, we would obviously get ap pizzas.
However, group of rotations divides pizzas into orbits. As it usually happens with
group action, order of orbit × order of stabilizer = order of group, which is p in our
case. Since p has only two divisors, we conclude that each pizza belongs to orbit of
1 or of p. Pizzas which belong to the orbit of 1 are monochrome pizzas (all
rotations keep preserve it, so all sectors are the same).
There are precisely a monochrome pizzas and ap – a non-monochrome states of
pizza, the second amount must be divided by p since it consists of p-orbits.
(b) Consider the following rotation of numbers: each number smaller than p is
increased by 1, and p is replaced by 1.
Like pizza rotations, this operation splits all subsets of p elements into orbits of p
or 1. There are only two stable sets of order p, which are invariant with respect to
that rotation: {1, 2, … , p} and {p+1, p+2, ... , 2p}.
Obviously, sums of elements in those two sets are divisible by p.
2p
The other    2 sets are divided into orbits of p. Consider one orbit. Sets in it
 p 
have k elements not bigger than p, where 0 < k < p. Each rotation adds k to the sum
of elements mod p. So, in each orbit of p there will be only one set with the sum
divisible by p.
3*. The order of group G is 2k(2l+1), and the group has an element of order 2k.
Prove that elements of odd order form a subgroup.
Solution. We shall use induction on k. for k = 0 the statement is obvious.
Consider left action of group on itself. Each element of a group, when you multiply
group elements by it, defines a permutation of group elements. Any element of odd
order defines only odd cycles, so it defines an even permutation. An element of
order 2k defines 2m+1 cycles of even order (2k), and that is odd permutation.
Consider a subgroup of those group elements that correspond to even
permutations.
Those elements contain all elements of odd order, but not all elements in the group.
It is a subgroup of order 2, so we reduced our problem to a problem on a smaller
group, which follows from induction assumption.
4**. Prove that the group of rotations of dodecahedron is A5 (the group of even
permutations of 5 elements), and groups of all dodecahedron symmetries is S5 (the
group of all permutation of 5 elements).
Solution. The statement about S5 is false. Dodecahedron happens to have a
transformation of central symmetry. Choose dodecahedrons center as a origin of
Cartesian system, central symmetry is multiplication by -1 and all other
transformation are linear, and multiplication by -1 commutes with any linear
transformation.
So, the group of symmetries of dodecahedron has a nontrivial center, while S5
doesn’t.
Now we shall prove that is the group of rotations of dodecahedron is A5.
Remark. A convenient way to draw a convex polytope on
a plane is to perform stereographic projection from the
center of some face.
We paint the edges of dodecahedron in 5 colors (see the
picture). When you look at any face, each edge has the
same color as the edge containing the opposite vertex of
this face and lying outside this face. In that way, any edge defines all the edges of
the same color.
Every symmetry of the dodecahedron defines some permutation of colors.
That gives us a mapping from group of rotations of dodecahedron to the S 5.
There are 60 rotations: a given face can be sent to every face in 5 ways.
There are also 60 even permutations in S5.
There are 24 possible cyclic orders of 5 elements; on the faces of our
dodecahedron we see 12 different cyclic orders. Therefore, all rotations correspond
to different permutations. Therefore the mapping that we constructed is injective.
It remains to prove that all the permutations we get are even.
Rotating of any face around its axis gives cycle of order 5, which is an even
permutation. Any other rotation can be decomposed into product of those simple
rotations, so it also corresponds to an even permutation. QED.
Remark. Another ways to formulate this group action:
a. Middle points of edges form 5 regular octahedrons. Rotations
permute those octahedrons.
b. It is possible to inscribe 5 cubes into dodecahedron, so that the vertexes of
the cubes are also vertexes of dodecahedron. Rotations permute those cubes.
I’ve decided that it is easiest to explain when picture can be made planar.
5. Let p be a prime number.
(a) Show that group of order pk has nontrivial center (in other words, it has
elements other than unit that commute with all the other elements).
(b) Show that any group of p2 elements is commutative.
(c) Can a group of p3 elements be non-commutative?
Solution. (a) Consider conjugation action of group by itself. That is, each element
g in the group G defines a permutation: h goes to ghg-1.
So, each element h has certain orbit: orbit is the set of all elements, to which h is
sent by different g. A very naïve, but also a very powerful fact: when group acts on
the set, the order of orbit × order of stabilizer (a subgroup of elements which don’t
move the element) = order of group.
In our specific case, order of group is pk, so any orbit is of size power of p. That
means, each orbit is either divisible by p or 1. Orbit of 1 under conjugation is of
size 1, so there are several more orbits of size one, since one orbit of size 1 +
several orbits of size divisible by p will never give you pk.
Elements having orbit of size 1 are central elements, so center is nontrivial.
(b) Consider center C of group G. By (a), it is nontrivial. So C has p or p2
elements. If it is p2, we won, so from now on we shall assume it is p.
Let a be an element outside C. Then a and C span a commutative group. Its order
is more than p, but still a divisor of p2, so it is p2. So it is the whole G and like we
said, it is commutative.
1 x y
(c) For example, consider matrixes over Z p of the form  0 1 z  .
0 0 1


It is easy to check they form a non-commutative group of p3 elements.