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.

Let and be algebraic number fields with , and having integer rings and , respectively. Suppose is an ideal such that is principal. Show that is principal in .

We will let and denote the ideal class groups of and , respectively. For brevity, if is a subset, then we will denote by the ideal generated by in and by the ideal generated by in .

We claim that if , then . To see this, suppose we have and such that . Then . Now if , then each is in , and so has the form . So . Conversely, , and so .

The mapping then induces a well-defined mapping given by . Moreover, since , is a group homomorphism.

In particular, if in , then in , as desired.

Consider the following commutative diagram of groups.

Suppose the rows are exact. That is, and for . Prove the following.

- If is surjective and and are injective, then is injective.
- If is injective and and are surjective, then is surjective.

- Let . Note that , so that . Since is injective, . Thus , and so . Say such that . Now , so that . Thus . Say with . Since is surjective, there exists such that . Now . Since is injective, we have . In particular, , so that . Thus . So , and so is injective.
- Let . Now . Since is surjective, there exists such that . Now , so that . Now , so that . Since is injective, . Thus . So . Say such that . Now . Thus , so that . Now ; say such that . Since is surjective, there exists such that . Now . Thus , and so . Thus is surjective.

Consider the following commutative diagram of groups.

Suppose the rows are exact; that is, for . Prove the following.

- If and are surjective and is injective, then is injective.
- If , , and are injective, then is injective.
- If , , and are surjective, then is surjective.
- If is surjective and and are injective, then is surjective.

- Let . Since is surjective, there exists such that . Now , so that , and so . Thus , and so . Say where . Since is surjective, there exists such that . Now , so that , and so . Since is injective, . In particular, , so that . Then . Thus , and so is injective.
- Let . Now , so that , so that . Thus . Since is injective, , so that . So . Thus there exists such that . Now , so that . Since and are injective, is injective, so that . Thus , and thus . Thus is injective.
- Let . Now . Since and are surjective, is surjective, so that there exists such that . Note that , so that . Thus , and so there exists such that . Since is surjective, there exists such that . Now . Thus , and so is surjective.
- Let . Now . Since is surjective, there exists such that . Now . Since is injective, . So , and so . Say such that . Now . Since is injective, . Thus is surjective.

Let be a ring with 1. Give an explicit example of a map from one -module to another which is a group homomorphism but which is not an -module homomorphism.

Let be any noncommutative ring, and fix such that is not in the center of . Consider as a left module over itself via left multiplication.

Note that the mapping given by is a group homomorphism since . However, if we choose not commuting with , then while . These two are not equal, so that is not an -module homomorphism.

Let be the ring of all continuous real valued functions on the closed interval . Prove that the map given by is a homomorphism of additive groups but not a ring homomorphism.

We know from calculus that . Thus is an additive group homomorphism.

On the other hand, note that , while . Thus is not a ring homomorphism.

Let and be groups, let be a homomorphism, and let . Identify and as subgroups of .

Prove that . (Recall that .)

Let . By definition, if , then . By the definition of multiplication in a semidirect product, we have , so that . Comparing first coordinates, . Since is arbitrary, we have , so that .

Suppose . Then . Now let be arbitrary. Then . Thus .

Let and be groups. Suppose and are subgroups and that there exists an isomorphism . Define by . Note that is normal in since , using a previous theorem. We denote by the quotient . (In particular, depends on .) Think of as the direct product “collapsed” by identifying each element with its -image in .

- Prove that the images of and in are isomorphic to and , respectively, and that these images intersect in a central subgroup isomorphic to . Find .
- Let . Let and as usual. Let be the central product of and which identifies and . (I.e. , , and .) Let be the central product of and which identifies and . (I.e. , , and .) Prove that .

- Let denote the canonical projection, and identify with and with in .
Let , and suppose . Then , so that . Thus , and hence is injective. Similarly, is injective.

Note that the restriction is also injective. Suppose ; then for some and we have . Thus ; by definition, then, and . Note that for all we have , so that is in the center of . Moreover, . Conversely, if , we have , so that . Then , and by the First Isomorphism Theorem, . In addition, is a central subgroup of .

Finally, by Lagrange write and . Note that . Now . We may also write this equation in the form .

- Define as follows: , , and . Because generates and these images satisfy the relations , , , and , extends to a homomorphism .
If we let denote the natural projection, is a mapping . We claim that is contained in the kernel of this mapping; it suffices to show this for the (unique) nonidentity element of , namely . In fact, = . Thus . By the remarks on page 100 in the text, there exists a unique group homomorphism such that , and in particular, .

We now show that . Suppose . Then ; note that for some integers , and that . Then .

If , then mod 2 and mod 4. Now if mod 4, then mod 4 and . If mod 4, then mod 2 and .

If , then mod 2 and mod 4. If mod 4, then mod 4 and . If mod 4, then mod 4 and .

Thus , hence is injective.

Now by part (a), we see that both and have order 16. Thus is an isomorphism.

Let be the additive group of real numbers and the multiplicative group of complex numbers with absolute value 1. Let be the homomorphism . Find the kernel of and the fibers over , , and .

Lemma: Let be set mappings such that exists, and let denote preimages. Then . Proof: If , then , so that . Thus , so that . If , then , so that , hence .

Note that , where is the mapping considered in the previous exercise and . Using the lemma, we see that , , , and .