## Every group is noncanonically isomorphic to its dual

For any group define the *dual group* (denoted ) to be the set of all homomorphisms into the multiplicative group of roots of unity in . Define a group operation in by pointwise multiplication of functions.

- Show that this operation on makes this set an abelian group. [Hint: show that the trivial homomorphism is the identity and that .]
- If is a finite abelian group, prove that . [Hint: write as and if , define to be the homomorphism which sends to and all other to 1. Then show that has order and that .]

This result is often stated as follows: a finite abelian group is self-dual. It implies that the subgroup lattice diagram of a finite abelian group is the same when turned upside down. Note however that there is no *natural* isomorphism ; the isomorphism depends on the choice of generators . This is often stated in the form: a finite abelian group is *noncanonically* isomorphic to its dual.

- Let be a group and an abelian group, and denote by the set of group homomorphisms . We define a product on pointwise. It is clear that this product is associative. Note that the trivial homomorphism satisfies and , so that is an identity under pointwise multiplication. Now let and define by . Note that , since is abelian. So . Moreover, for all , so that . Hence is a group. Finally, note that , so that is abelian.
- Define each as in the problem statement. Since (by this previous exercise) each is a homomorphism, for all .
Now let . We have . Thus . By this previous exercise, there exists a unique group homomorphism such that .

Note that each homomorphism is uniquely determined by its values at the (by this previous exercise), and that is a th root of unity. Thus for some (since is a “primitive th root of unity”, whatever that means.). Consider now ; note that since is a homomorphism, and that this equals . By uniqueness, then, and because is arbitrary, . Thus is surjective.

Suppose now that . Then by definition . Now let ; we have . Thus divides , hence mod . Thus we have , so the kernel of is trivial and thus is injective.

Hence .

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