But if, in a finite field, for some non-zero element of the field, then we can show that every element of the field is a sum of two cubes (Problem 2).

**Problem 1**. Show that every element of a finite field is a sum of two squares.

**Solution**. Let be a finite field. So we want to show that if then for some We can actually be more specific if we consider two cases. Let

Case 1: for some integer Then, since for all we get So in this case, every element of the field is a square.

Case 2: for some integer Since is finite, the multiplicative group is cyclic.

So Let and consider the sets

Clearly and Thus and hence

Therefore i.e. there exist such that and the result follows.

**Remark 1**. Regarding the second case in the solution of Problem 1, notice that, in fact, we have

and so The reason is that if for some integers then and hence must be divisible by implying that is even.

**Problem 2**. Let be a finite field and suppose that there exists such that Show that every element of is a sum of two cubes.

**Solution**. So we want to show that if then for some Let and let’s consider three cases.

Case 1: for some integer Then for all

Case 2: for some integer Then for all and clearly

So, in both cases 1 and 2, for every there exists such that

Case 3: for some integer Since is finite, the multiplicative group is cyclic. So Let and consider the sets

Clearly and So and

So at least two of the sets have non-empty intersection. If or then for some and we are done.

Now suppose that So there exist such that and so Since, as given in the problem, for some we have and Hence

**Remark 2**. Regarding the third case in the solution of Problem 2, notice that, in fact, we have

and so The reason is that if for some integers then and hence must be divisible by implying that is divisible by

]]>**Solution**. If or then gives or and we are done. Otherwise, there exist and But then contradiction!

So, as a result, if is a finite group and are two subgroups of with and then That raises this question: how large could get? The following problem answers this question.

**Problem 2**. Let be a finite group and suppose that are two subgroups of such that and Show that

**Solution**. Recall that and thus Hence and so

where and

Now, since and we have and i.e. and So if we let and then and thus

The result now follows from

**Example 1**. The upper bound in Problem 2 cannot be improved, i.e. there exists a group and subgroups of such that An example is the Klein-four group and the subgroups and Then and

**Example 2**. We showed in Problem 1 that a group can never be equal to the union of two of its proper subgroups. But there are groups that are equal to the union of three of their proper subgroups. The smallest example, again, is the Klein-four group

Let be a group and let If are conjugate in i.e. for some then clearly if and only if So have the same order. The converse however is false, i.e. if have the same order, that does not imply are conjugate. For example, in an abelian group, two elements are conjugate if and only if they are equal but you can obviously have distinct elements of the same order in the group, e.g. in both non-zero elements have the same order

We are now going to show that although two elements of the same order of a group might not be conjugate in the group, but they are certainly conjugate in some larger group.

**Problem**. Let be a group and suppose that have the same order. Show that there exists a group such that are conjugate in

**Solution**. By Cayley’s theorem, we can embed into the symmetric group using the injective group homomorphism defined by where is the permutation defined by for all So we only need to show that are conjugate in Well, let Then the cycle decomposition of are in the form

and

So have the same cycle type and hence they are conjugate in

]]>But what if doesn’t have The following problem answers this question.

**Problem**. Let be a ring, which may or may not have Show that if has no proper left ideals, then either is a division ring or and for some prime number

**Solution**. Let

Then is a left ideal of because it’s clearly a subgroup of and, for and we have and so i.e. So either or

Case 1: That means for all or, equivalently, Thus every subgroup of is a left (in fact, two-sided) ideal of Hence has no proper subgroup (because has no proper left ideals) and thus for some prime

Case 2: Choose So and hence because is clearly a left ideal of Thus there exists such that Now

the left-annihilator of in , is obviously a left ideal of and we can’t have because then So Since

we have Thus Let

Clearly is a left ideal of and Thus So for all Now let be any element of Then, by what we just proved, On the other hand, by the same argument we used for we find such that Thus i.e.

So and hence i.e. and thus

So and hence for all Thus proving that is a division ring.

**Remark**. The same result given in the above problem holds if has no proper right ideals.

**Example**. Let be a prime number. The ring

is not a division ring and it has no proper left (or right) ideals.

]]>In other words, is -abelian if the map defined by is a group homomorphism.

**Definition 2**. If is both -abelian and -abelian, for some integers then is also -abelian because then for we will have

So the set

i.e., the set of those integers for which is -abelian, is a multiplicative subset of Clearly The set is called the **exponent semigroup** of

**Remark 1**. Since we have if and only if So a group is -abelian if and only if or, equivalently, for all So a group is -abelian if and only if it is -abelian. In other words, if and only if

**Example 1**. Every abelian group is obviously -abelian for all It is also clear that -abelian groups are abelian because gives As the next two examples show, there exists a non-abelian -abelian group for any

**Example 2**. Let be an odd integer and consider the Heisenberg group which is a non-abelian group (why?). We show that i.e. is both -abelian and -abelian. To see that, let

An easy induction shows that

for all integers Thus the identity matrix, and so for all

Hence and for all proving that is both -abelian and -abelian.

**Remark 2**. Notice that, in Example 2, if is even, then (as elements of and so is not always the identity element in this case. However, there’s a way to fix this, as the next example shows. But first, notice that can also be viewed as the group of triples

with multiplication defined by

In the next example, we modify the above multiplication such that we get for all

**Example 3**. Let be any integer and consider the set Define multiplication in by

It’s easy to see that is a non-abelian group. We show that Let A quick induction shows that

for all integers So for all and thus for all proving that is -abelian.

Also, since for all we have and so is -abelian.

**Problem 1**. Let be a group. Show that

i) if for some integer then for all

ii) if for some integer then is abelian.

**Solution**. i) Let We have

and so

ii) Let By i), we have and Thus

and so

**Problem 2**. Let be an -abelian group and let Show that

i)

ii)

**Solution**. i) By Remark 1, for all Thus

and the result follows.

ii) Again, using the Remark 1, we have

**Remark 3**. Let be an -abelian group for some integer An immediate result of Problem 2, ii), is that if either is torsion-free or is finite and then is abelian.

In this post, we showed that the ring has idempotents, where is the number of prime divisors of Clearly is also the number of prime (= maximal) ideals of (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let be the Euler’s totient function. The number of units of is If is the prime factorization of then If is odd, then is even for all and hence divides So if is odd, then the number of idempotents of divides the number of units of As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring is called **semilocal** if the number of maximal ideals of is finite and it is called **local** if it has only one maximal ideal.

**Example 1**. If is a prime and is an integer, then is a local ring with the unique maximal ideal If is an integer, then is a semilocal ring (what are its maximal ideals?).

**Example 2**. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let be an Artinian ring and let be the set of all finite intersections of maximal ideals of Since is Artinian, has a minimal element Let be any maximal ideal of Since and is a minimal element of we must have So for some and hence Therefore are all the maximal ideals of

**Problem 1**. Show that if is a local ring, then are the only idempotents of

**Solution**. Let be the maximal ideal of and let be an idempotent of Then Thus either or If then because otherwise which is false. So is a unit because there’s no other maximal ideal to contain So for some and thus Similarly, if then and thus is a unit. So for some implying that and hence

**Problem 2 **Show that the number of idempotents of a commutative Arinian ring is where is the number of maximal ideals of

**Solution**. Since is Artinian, it has only finitely many maximal ideals, say (see Example 2). The Jacobson radical of is nilpotent, hence there exists an integer such that

Thus, by the Chinese remainder theorem for commutative rings, Since each is a local ring, with the unique maximal ideal it has only two idempotents, by Problem 1, and so has exactly idempotents.

**Problem 3**. Let be a finite commutative ring. Show that if the number of elements of is odd, then the number of idempotents of divides the number of units of

**Solution**. Since is finite, it is Artinian. Let be the set of maximal ideals of By problem 2, the umber of idempotents of is and for some integer

Since is odd, each is odd too because is a subgroup of and so divides Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each is which is an even number because both and are odd. So the number of units of which is the product of the number of units of is divisible by which is the number of idempotents of

**Remark 1**. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, has one unit and two idempotents. However, the result is true in which has units and two idempotents.

Can we find all even integers for which the result given in Problem 3 is true in ? Probably not because this question is equivalent to finding all integers such that where is the number of prime divisors of and that is not an easy thing to do.

**Remark 2**. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider the ring of matrices with entries from the field of order three. Then has units (see Problem 3 in this post!) but, according to my calculations, has idempotents and does not divide

**Solution**. Suppose first that is a division ring. Then, since is a finite ring, is a finite field, by the Wedderburn’s little theorem. Let Then for all and so has no solution in

Conversely, suppose that is not a division ring (equivalently, a field because is finite). So and hence the equation has solutions in (just choose and ).

Let be the Jacobson radical of Since is finite, it is Artinian and so is nilpotent.

So if then there exists such that and so for all Therefore the equation has a solution in

If then by the Artin-Wedderburn’s theorem,

for some finite fields Since is not a field, we have either for some or for all and If for some then and hence will have a non-zero nilpotent element and we are done. If for all and then

will satisfy

]]>**Solution**. The proof is by induction over There’s nothing to prove for For since we have

and thus, by the induction hypothesis, for all So for some

If then and so Hence, by the induction hypothesis, for all

If then (so because ). Thus

and so, by the induction hypothesis, for all

If then gives and so implying, again by the induction hypothesis,that for all

**Example**. Show that for all integers

**Solution**. Let where for all Since and for all we have, by the above problem, for all Thus and we are done because obviously

It is known that is invertible and if then We are going to use these two properties of Hilbert matrices to solve the following calculus problem.

**Problem**. Let be an integer and let be a continuous function. Suppose that for all Show that

**Solution**. Since the Hilbert matrix, is invertible, there exist real numbers such that

So the polynomial satisfies the conditions

Clearly is the sum of all the entries of and so Now let be a real-valued continuous function on such that

Let be the above polynomial.Then since

integrating gives

]]>**Solution**. Since is orthogonal, its eigenvalues have absolute value and it can be be diagonalized. Let be a diagonal matrix such that for some invertible matrix Then

We claim that the eigenvalues of are for some Well, the characteristic polynomial of has degree three and so it has either three real roots or only one real root. Also, the complex conjugate of a root of a polynomial with real coefficients is also a root. So, since the eigenvalues of are either all which is the case or two of them are and one is which is the case or one is and the other two are in the form for some So

Note that given the matrix

is orthogonal, and

]]>