Let be a field and be a finite multiplicative subgroup. Prove that is cyclic.

We prove this first in the case that is a -group, say , by induction on . Given , we let denote the number of elements of of order .

Suppose ; then certainly is cyclic and for each dividing .

Suppose that every subgroup of order is cyclic for some and that for all such , . Let be a subgroup of order . Note that the elements of are precisely the roots of in . Similarly, the roots of are precisely those elements of whose order divides . Suppose and are two such roots; that is, and . Now , so that . Since , the set of roots of (i.e. the set of elements of whose order divides ) form a subgroup of of order . By the induction hypothesis, this group is cyclic, and for each . Note also that for each such , with respect to is equal to with respect to this subgroup. Since , we have . In particular, contains an element of order , and so is cyclic.

Thus every finite -subgroup of is cyclic. Now let be an arbitrary finite subgroup; since is abelian, it is the direct product of its -subgroups. Since these are cyclic, is cyclic.

Let be a prime. Let and be cyclic groups of order with generators and , respectively. Set so that is the elementary abelian group of order . Prove that the distinct subgroups of of order are , , …, , .

We saw previously (by counting) that has precisely distinct subgroups of order . Thus it suffices to show that these subgroups are mutually distinct.

Suppose . Then for some , hence . Thus , so that mod . Likewise, mod .

Suppose . Then since otherwise the first coordinate of each element of is 1, a contradiction since . Thus none of the first subgroups above is equal to the last. If (without loss of generality, considering the generators chosen above) , then , and we have mod . Thus the first subgroups in the list above are mutually distinct. Hence the subgroups given above exhaust the distinct order -subgroups of .

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 all subgroups of are normal, where is the elementary abelian group of order .

Let be a subgroup, and let and . Since is abelian, we have .

We saw previously that the conjugacy classes of are , , , , and . In particular, we have .

If , then . If , then (since every element of has order 2) . Thus is normal.

Let be groups and let . Let be a proper nonempty subset of and let . Define to be the set of elements in such that for all .

- Prove that is isomorphic to .
- Prove that is a normal subgroup of and that .
- Prove that .

- Define a mapping by . We need to show that is a bijective homomorphism.
- Homomorphism: Let . Then for each , . Hence , so that is a homomorphism.
- Surjective: Let . Define by if and otherwise. Clearly this is indeed an element of . Moreover, . Because is arbitrary, is surjective.
- Injective: Let . Suppose . By definition, for all . Moreover, for , we have . Hence , so that is injective.

- Let and . Note that . For , we have , so that . Thus . Hence is normal.
Now define a mapping as follows: if and otherwise. We wish to show that is a surjective homomorphism and that .

- Homomorphism: Let . If , then . If , then . Thus is a homomorphism.
- Surjective: It is clear that if , then .
- Kernel: Let . Then for all ; in particular, if . Thus . Let . Now if , . Clearly then , so that .

By the First Isomorphism Theorem, we have .

- Define a mapping as follows. for and otherwise, and for and otherwise. We wish to show that is a bijective homomorphism.
- Homomorphism: Let . If , then . If , then . Thus . Similarly, . Hence , so that is a homomorphism.
- Surjective: Let , with and . Define as follows: If then , and if then . Then by definition, we have if and otherwise, so that . Similarly, . Hence , and is surjective.
- Injective: Suppose such that . For , we have , and for , we have . Thus , and is injective.