Fix an matrix over . Define a mapping by . Prove that is a group homomorphism on .

Note that is nonsingular by this previous exercise, so that is properly defined.

Now by this previous exercise, which equals . So is a group homomorphism.

unnecessary lemmas. very sloppy. handwriting needs improvement.

Exhibit a set of representatives of the conjugacy classes of .

It suffices for us to give the (essentially unique) rational canonical matrix in each conjugacy class. If is a matrix over , then the minimal polynomial of has degree between 1 and 3. There are 14 such polynomials, which we list with their factorizations. (We showed which monic polynomials of degree at most 3 were irreducible in this previous exercise.)

- (irreducible)
- (irreducible)
- (irreducible)
- (irreducible)
- (irreducible)

Note that if divides the minimal polynomial of , then in particular is a zero divisor in and so cannot be a unit. So matrices in cannot have as a divisor of their minimal polynomial. This eliminates (1), (3), (5), (7), (9), (11), and (13) from our list of candidate minimal polynomials.

An irreducible polynomial of degree 2 is also unsuitable as a minimal polynomial, since the characteristic polynomial of (having degree 3) cannot divide any power of an irreducible polynomial of degree 2. (Either the characteristic polynomial is irreducible (bad) or has an irreducible linear factor (also bad).) This eliminates (6) from our list of candidate minimal polynomials. The six remaining candidates, once chosen as a minimal polynomial, determine the remaining invariant factors. The possible lists of invariant factors for are thus as follows.

- , ,
- ,

The corresponding rational canonical matrices are as follows.

These six matrices are a complete and nonredundant set of representatives of the conjugacy classes of .

Compute, up to similarity, all the elements of having order 4. Do the same in .

Our task is to find all the matrices over which satisfy but not or . Note that over , . If is a matrix of order 4, then the minimal polynomial must be divisible by and must divide . Since the characteristic polynomial of has degree 2, there is only one possible list of invariant factors for , namely . The corresponding rational canonical form matrix is . Thus every element of of order 4 is similar to this matrix.

Now over , we have . If is a matrix of order 4, then its minimal polynomial must be divisible by either or , must divide , and must have degree at most 2. There are seven possibilities for the minimal polynomial of , and with the minimal polynomial chosen, the remaining invariant factors are determined (since the characteristic polynomial has degree 2). The possible lists of invariant factors are as follows.

- ,
- ,

The corresponding rational canonical forms are as follows.

Every matrix in of order 4 is similar (i.e. conjugate) to exactly one matrix in this list.

Let be an odd prime. Prove that every element of order 2 in is conjugate to a diagonal matrix with s on the diagonal. Classify the groups of order .

We begin with some lemmas.

Lemma 1: Let be a field in which . Every element of order 2 in is conjugate to a diagonal matrix. Proof: Let have order 2. Now ; Suppose . Comparing entries, we see that , so that . If , then since , . Similarly, if , then . Thus either or . If , then is already diagonal. Suppose .

- If and , let . Evidently and is diagonal.
- If and , let . Evidently and is diagonal.
- If and , let . Evidently and is diagonal.

Lemma 2: If is a diagonal matrix of order 2, then and . Proof: We have .

Lemma 3: Let be an odd prime. Let be integers such that for all positive integers , mod . Then mod and mod 2. Proof: If mod , let . Then mod for all , which is absurd. Thus mod . Now suppose mod ; we have mod . Thus mod 2.

Now to the main result.

Let be a group of order , where is an odd prime. Let be a Sylow -subgroup. Since , is normal in . Let be a Sylow 2-subgroup. Since , we have . By the recognition theorem for semidirect products, for some group homomorphism . Evidently, classifying groups of order is equivalently to finding the distinct semidirect products of this type.

Let .

Let . Now ; say , where . Note that ; thus there is a unique element of order 2 in ; namely . Thus there are exactly two homomorphisms . yields the semidirect product , and yields the nonabelian group . These are distinct since one is abelian and one is not.

Now let . Now . Let , , , and . Note that if , then has order 1 or 2. If is 1 or 2, then is conjugate to for some by Lemmas 1 and 2. By a previous theorem, then, the distinct semidirect products of this form are isomorphic to one of . Let , , and . Note that ; by a lemma to a previous exercise, .

We claim that , , and are nonisomorphic. Now is abelian while the other two are not, so is distinct from and .

We will now find presentations for and .

Note that is generated by , , and , and . If we write , then . Similarly, . Thus has the presentation .

is also generated by , , and , and . Similarly, and . Thus has the presentation .

We will now compute the center of . We can see that every element of has the form where and . There are such forms, and we know by other means that . Thus every element can be written in this form uniquely. Let and ; using several previous lemmas, we have that is equal to . Comparing exponents and using Lemma 3, we have mod and mod 2; thus , and we have .

Note that is nontrivial since . In particular, .

Thus there are five distinct groups of order , an odd prime.

Let , where is a field, be the group of upper triangular matrices all of whose diagonal entries are equal. Prove that , where is the group of nonzero multiples of the identity matrix and the group of strictly upper triangular matrices.

Recall that is normal in , where denotes the upper triangular matrices with only 1 on the diagonal. Then is normal in . Now if , we have for some nonzero field element . Then for all , we have , so that ; more specifically, is normal in .

We can see that is trivial, since every element in has only 1s on the main diagonal. Finally, if has the element on the main diagonal, then , and in fact . Thus ; by the recognition theorem, we have .

Let be a field for all and use the preceding exercise to show that the set of matrices over such that each row and column contains exactly one 1 and all other entries are 0 is a subgroup of isomorphic to . (These matrices are called *permutation matrices* since they simply permute the standard basis (as above) of the -dimensional vector space .)

We know that if is a finite field then . This isomorphism can be defined as follows: given , is the matrix in whose th row is precisely . (In particular, is not canonical since it depends on the choice and order of a basis; here we choose the standard basis.) In the previous exercises, we found an injective group homomorphism . Combining these results we have an injective group homomorphism , computed as follows: . We can see that each is obtained from the identity matrix by permuting the rows, so that each has a single 1 in each row and column and 0 in all other entries. Thus is identified with a subgroup of consisting of permutation matrices; moreover, by counting we see that all permutation matrices are represented this way.

Thus we have proven the result for a finite field ; we began with this case because it was shown previously that . If this is true for arbitrary fields then the same proof carries over to arbitrary fields ; however this will not be proven until later in the text. In the meantime we can convince ourselves that the result holds over arbitrary fields by noting that, in computing the product of two permutation matrices, we never deal with numbers other than 1 or 0. In particular, the question of whether or not in for some integer never arises, so that all computations hold for arbitrary fields (in which ) and in fact the set of permutation matrices over any is closed under matrix multiplication.

Prove that the number of Sylow -subgroups of is .

We know that . By Sylow’s Theorem, the number of Sylow -subgroups divides and is congruent to 1 mod .

We saw in the previous exercise that the strictly upper triangular matrices form a Sylow -subgroup. We now show that .

Let be an invertible upper triangular matrix; then . Now if is strictly upper triangular, we have , which is strictly upper triangular. Thus the normalizer of contains the upper triangular matrices in . There are of these, since diagonal entries are arbitrary nonzero field elements, while the upper right entry is unrestricted. Now . By Sylow’s Theorem, then, divides . Since is congruent to 1 mod , either or ; however, since the strictly upper and strictly lower triangular matrices are distinct Sylow -subgroups of .

Show that the subgroup of strictly upper triangular matrices (i.e. with only 1 in diagonal entries) in is a Sylow -subgroup.

We know that the order of is . Note that does not divide for any . Moreover, we have . Thus (by definition) a Sylow -subgroup of has elements.

Recall that a matrix is strictly upper triangular if all diagonal entries are 1, all lower entries are 0, and all upper entries can be any field element. Then the number of elements in is , where is the number of entries above the main diagonal. We can see that this number is . Thus is a Sylow -subgroup in .

Let and let . Prove that every element of is conjugate to some element of . Deduce that is the union of conjugates of . [Hint: Show that every element of has an eigenvector.]

Let ; then .

- If , then .
- If , let . Note that , so that ; moreover, . Then .
- If and , let . Note that , so that . Evidently . Note that .
- Suppose now that and . Let , where . Note that , so that . Evidently . Moreover, we see that .

Thus every element of is conjugate to some element of ; clearly then .

Let be a finite field of order and let be a positive integer. Prove that .

We proved in a previous exercise that , where denotes the set of all elements in which have a multiplicative inverse. By definition, this is all elements except 0. So the index is .