We first show that it is sufficient to construct a weighted directed graph having the given group as automorphism group. A weighted directed graph is a triple , where is a finite set of vertices, is a finite set of weights and is a partial map. An automorphism of is a bijection such that for all , is defined if and only if is defined in which case . The group of all automorphisms of is denoted by . If , then denotes the (total) degree of .

Lemma. Let be a weighted directed graph with for all . Then there exists a simple graph with .
Proof. Wlog assume that for some . Construct from by replacing every edge with by

and notice that every automorphism of induces an automorphism of in the canonical way while it is easy to see that every automorphism of must necessarily permute the original vertices and thus be equal to one of the induced automorphisms.

Let be an arbitrary finite group with order at least three (the other cases are trivial). Construct a weighted directed graph by putting and for all , and for , let . Note that is an automorphism of the graph for all since and that .
Furthermore, if is an automorphism of the graph, let and be arbitrary. Then , so , i.e. . It follows that and by the lemma above, there exists also a simple graph with automorphism group .

is well known and easy to prove. One can now ask what happens if the sum is taken only over the primitive th roots of unity.
Let

be the sum of all primitive th roots of unity. Since every th root of unity is a primitive th root of unity for some divisor of , we have, for all positive integers ,

,

This property however is the characterisation of the Möbius function (namely being the Dirichlet inverse wrt the constant function), hence, .

Problem. Choose two natural numbers randomly (with uniform probability) and independently. Let be the probability that and are coprime. Find .

The result is but before proving this rigorously, I will present two heuristic approaches here.

Heuristic 1. Let be a prime number. Then intuitively, the probability that two randomly chosen integers are not both divisible by is . Thus, the probability that two randomly chosen integers are neither both divisible by nor by etc is

The product above is exactly and it is natural to conjecture that the remaining terms are negligible (although the direct proof of this might be quite nasty).

Heuristic 2. Suppose that is the probability in question. Then for a positive integer , the probability that two randomly chosen positive integers have is since is the probability that both numbers are divisible by and is the probability that the remaining factors are coprime. But the sum of all those probabilities must be equal to (as any two randomly chosen positive integers have a ), so

A general problem with these heuristic approaches is among others that there is nothing like a uniform probabilistic distribution over . The second approach seems rather more likely to be rigorizable, I will however give a proof with a completely different approach here.

Proof. First, notice that

where denotes the Euler’s totient function. Recall that

where denotes the Moebius function, so

 
       
                    
                  
                   
          .

Recall that for all real numbers ,

Thus,

as required.

Recall that

so for all .

Denote by the polynomial defined above. We know that , but can we find the exact value? For this purpose, consider the following very useful lemma.

Lemma. Let be an integral domain and be a finite multiplicative subgroup of with . Then for every integer ,

Proof. It is immediate from Lagrange’s Theorem that if . Suppose that . Notice that must necessarily be cyclic (as e.g. can easily be proved using the structure theorem of finitely generated Abelian groups) so in particular, there exists a with . Multiplication with permutes the elements of , so

But and is an integral domain, so the result follows.

Corollary. If is a finite field with elements and is an integer, then

Suppose now that

where . For integers , consider now the sum

                    

But

We thus obtain

Theorem. Suppose that the coefficient of in is nonzero. Then if and only if

for all with for all , and

where the latter sum must then be equal to .

Theorem. Let be a graph and let be the set of all vertices of even degree. Then for each , the number of Hamiltonian paths from to some vertex in is even.

Proof. For a Hamiltonian path and a vertex with , let be the Hamiltonian path . We call the transformation the path exchange of wrt . Apparently, if , then conversely .

For a fixed , let be the set of all Hamiltonian paths starting in . Draw a graph (called the -path graph of ) with vertices , where two Hamiltonian paths are connected in iff for some . For each , the degree of in is , where is the end vertex of other than . In particular, and are of different parity. But in every graph, the number of vertices of odd degree is even. This proves the claim.

Corollary. Let be a graph with being odd for all . Then for each , the number of Hamiltonian cycles passing through is even. In particular, if in addition is Hamiltonian, then it has at least two distinct Hamiltonian cycles.

Remark. The special case of the above corollary for cubic graphs is known as Smith’s Theorem. However, the problem of finding a second Hamiltonian cycle (in polynomial time) in a cubic graph when given already one is still an open question in complexity theory.

Theorem (Combinatorial Nullstellensatz). Let be an arbitrary field and be a polynomial with , where are nonnegative integers such that coefficient of in is nonzero. Suppose that are subsets of with for all . Then there exist so that .

This theorem immediately implies that if is a nonconstant polynomial annihilating , then for some (there are of course other nice proofs of that result but I somehow like it how CNS trivialises it). Furthermore, we can infer from this that contains exactly elements when considered as functions . Indeed, there are at most polynomial functions since there are exactly functions and two distinct polynomials with for all (of which there are exactly ) cannot be equal on the whole of by the above. We therfore immediately see that every function is equal to a polynomial function.

Also, an immediate inductions on the degree of the polynomials show that the Ideal of polynomials annihilating is generated by the polynomials .

I will prove the following result:

Proposition. Let be a positive integer. Then every finite group of order is cyclic if and only if

where denotes the Euler’s Totient Function.

I will use the following notation: If is a group and , then denotes the centraliser of wrt to and denotes the cyclic group generated by .

Proof of the if-part. Let be a positive integer satisfying .
Note that using the canonical formula for , it immediately follows that is squarefree.

We will use induction on . The claim is trivially true if is a prime number. Suppose now that for all positive integers with , all groups of order are cyclic.

Let be a group of order . We have to prove that is cyclic. From the induction hypothesis and Lagrange’s theorem, it follows that all proper subgroups of are cyclic.

Lemma 1. Any two different elements of a cyclic subgroup of are not conjugate.
Proof. Let be a cyclic subgroup of . Suppose that for integers and some . We claim that .
An immediate induction on shows that

holds for all positive integers . Putting , we see that

,

so divides . Clearly, and divide . Furthermore, is squarefree.

Let be a prime divisor of . Then if and only if . If and , then clearly .
Otherwise, if and are not divisible by , then has a multiplicative inverse modulo , so . Let . Then , so . Also, since by Fermat’s little theorem. But , so . It follows that is a common divisor of and , so . Hence, , so .

Thus, we see that for all prime divisors of and since is squarefree, it follows that . But then , as required.

Next, we prove that the center of is not trivial. Hence, assume for the sake of contradiction that the center of is trivial.

Then for each with , is a proper subgroup of . Furthermore, this subgroup is maximal, for if is a maximal proper subgroup containing (we can assume it to be cyclic by the induction hypothesis), then , so .

It follows that

Lemma 2. For each with , is the unique maximal proper subgroup of containing .

It is well known that if is a cylic group of order and is a divisor of then contains a unique subgroup of order , namely .

Lemma 3. Let and be two maximal proper subgroups of and let and . Suppose that . Then and are conjugate. In particular, .
Proof. Let be a common prime divisor of and . Then and both contain a unique subgroup and of order respectively. It follows from Lemma 2 that and .

Furthermore, it follows from Sylow’s second theorem that and are conjugate, so for some . Thus, is a generator of , wlog assume that . Now,

.

It follows that , so .

Now, since the center of is trivial, follows from the class equation that

,      (1)

where are representatives of the different nontrivial conjugacy classes. Each of the are maximal proper subgroups, the order of each two of them being either equal or coprime (Lemma 3). Furthermore, it is clear that for every prime divisor of , there exists an so that divides . Hence, we can write as , where each of the is the order of some of the . Notice that and for all . Every summand in (1) is of the form for some . Furthermore, for each , each of the elements of belong to different conjugacy classes by Lemma 1. Moreover, for each with , we have that since is a maximal proper subgroup of containing , which is unique by Lemma 2. It follows that in the sum in (1), the summand appears at least times for all . Hence,

so

Thus,

and since for all ,

so , which is the desired contradiction.

It thus follows that the center of is not trivial. Let and . Then and , so by the induction hypothesis, both and are cyclic. Let be a generator of and be a generator of . Let . Then , so . Let . Then .

Assume that . Then divides for some prime divisor of . Since ,

But and is squarefree, so is divisible by exactly one of the numbers and . But then, either

    or    

which is a contradiction. Hence , so is cyclic, as required.

Proof of the only if-part. Let be a positive integer satisfying . We have to show that there exists a finite group of order that is not cyclic. The result is obvious if is not squarefree, because if is a multiple prime divisor of , then the group is not cyclic.

Suppose now that is squarefree and . Then there exist prime divisors and of so that is divisible by .

Let be a primitive root modulo . We consider the set

Clearly, since for all if and only if and .

Furthermore, is a subgroup of the symmetric group (the subgroup conditions can readily be verified). Let send to and send to . Then sends to and sends to , so is not abelian.

Consider now the group . Since , . But is not abelian, so is not abelian. In particular, is not cyclic. This completes the proof.

Remark. For , the group constructed in the proof of the “only if”-part above is isomorphic to the dihedral group of a regular -gon.

Further Remark. I have just noticed that the group is acutally isomorphic to the semidirect product wrt to the homomorphism

,
.

Some time ago, I found the following problem on a problem sheet.

Problem. Let and be matrices over . Prove that there exist complex matrices such that

if and only if there does not exist a vector with , where is the identity matrix.

One cannot help noticing the apparent relation of this problem to Bezout’s Lemma which however can obviously not be applied directly as is not commutative (and thus not a principal ideal ring) but we will see that commutativity is acutally not required for Bezout. In fact, it is sufficient that every finitely generated left ideal is principal (or rather every right ideal, if the linear combinations in question are supposed to be from the right).

In course of thinking about the problem above, I discovered that in , every finitely generated left ideal is principal and that the generating elements are actually “easy” to describe. Hereby, is a vector space over an arbitrary field and denotes the set of all -endomorphisms.

I must at this point admit that my knowledge in linear algebra is actually quite moderate (I’m not at university yet). If someone sees things in a broader light and recognises these results as well known or as special cases/corollaries of more general results (I’m quite sure they are), I’d be glad about feedback.

Theorem. Let be a (not necessarily finite-dimensional) vector space over a field and let

be an left ideal in , generated by . Then is principal and a generating element of can be described as follows. Define

.

Then any with (e.g. a projection on a complement of along ) generates .

Proof. First, it is easy to see that using induction, it is sufficient to prove just the case for .

Let be any -endomorphism with . Let be a base of which extends to a base of and a base of . Let and . Clearly, and extends to a base of , where . Then is a base of and is a family of linear independent vectors. Let and be complements of and to respectively.

We now can define the -endomorphisms by for and for all , and by for all and for all . Note that and are well defined. Now, let . Then

                ,

so . It follows that .

In order to prove that indeed generates , it is sufficient to prove the following result: if and , then there exists a linear function such that .

Let be a base of . This extends to a base of and to a base of . Then is a base of , which has a complement to . But then we can simply define a linear function by for and for . Note that is well defined. But then, for all since for all . Thus, .

Remark. In the case when is finite-dimensional over , we can prove similar results for right ideals by using the transposes of linear maps. Moreover, it is easy to see that if is finite-dimensional, then every left ideal of is principal and generated by a -endomorphism which minimises . Indeed, if is a left ideal of , is such a -endomorphism which minimises and is another arbitrary -endomorphism, then the subideal of generated by and is finitely generated, and thus generated by a single -endomorphism satisfying . But since is minimal, it follows that , implying . Hence, there exists a -endomorphism such that , so generates .

We therefore see that if , then is, in some sense, a non-commutative principal ideal ring.

Corollary. Let be a vector space over and let . Then there exist with

if and only if .

Furthermore, if is finite-dimensional, then there exist such that

if and only if , where denotes the transpose of .

Some days ago, I was told that every field has an algebraic closure which is unique except of isomorphism (apparently, this seems to be a well known result). However, I wasn’t able to prove these results without using Zorn’s Lemma (neither the existence nor the uniqueness).

Definition. Let be a field. An Algebraic Closure of is a field so that is an algebraic field extension and is algebraically closed, i.e. every polynomial with coefficients in splits into linear factors in .

We first prove with Zorn’s Lemma that every field has an algebraic closure. Note that the union of the fields in a chain (ordered by set inclusion) is itself a field.

Theorem. Every field has an algebraic closure.
Proof. Let be the set of all fields so that is an algebraic field extension. Then is partially ordered by set inclusion.

Let be a chain in . Consider . Then clearly, is a field containing .

Moreover, suppose that . Then there exists some so that . Hence, is algebraic over as is an algebraic extension of . We therefore see that is an algebraic extension.

It follows that every chain in has an upper bound in , so has a maximal element by Zorn’s Lemma. We shall prove that is the required algebraic closure.

Suppose then that is nonconstant and irreducible. Let be a root of in some splitting field and let . Then is a finite and hence algebraic field extension. But is also an algebraic field extension, so is algebraic, contradicting the maximality of .

Next, we shall prove that the algebraic closure is unique up to isomorphism. We will use the following (well-known) Lemma.

Lemma. Let be an isomorphism between fields. Let be a polynomial with coefficients in and be the polynomial obtained by applying to the coefficients of . Let and be splitting fields of and over and respectively. Then there is an isomorphism extending .

Theorem. Let and be algebraic closures of a field . Then .
Proof. Consider the set of isomorphisms , where and . Define a partial order on where iff extends . Denote by and the domain and image of respectively.

Suppose that is a chain in . Define a function , for some with .

Note that is well-defined, since is a chain wrt . Furthermore, and are both fields as they are the union of chains of fields (wrt set inclusion) respectively, so is a homomorphism between fields. Since is non-trivial, it is injective. Moreover, for each , there is some with , so is surjective. It follows that is an upper bound of .

It follows from Zorn’s Lemma that that has a maximal element . We shall prove that is an isomorphism between and . Let and . Assume that and is some element in . Clearly, is algebraic over and thus over . Suppose is the minimal polynomial of over and be the polynomial obtained from by applying to the coefficients of and let and be splitting fields of and over and respectively. Then and . But by the Lemma above, there exists an isomorphism from to extending , which contradicts the maximality of . Thus, . Using the same argument for , we see that . Hence, , as required.

Remark. It is easy to see that the algebraic closure of a field can also be defined as the smallest algebraically closed extension field of , i.e. the intersection of all algebraically closed field extensions of (which, itself is an algebraically closed field extension of ).
Indeed, let be the algebraic closure of and let be the smallest algebraically closed field extension of . Then clearly, . Hence, every is algebraic over , so is an algebraic field extension. But then, is an algebraic closure and we have already seen that algebraic closures are unique up to isomorphism. It follows that