Every group is noncanonically isomorphic to its dual

For any group G define the dual group G (denoted \check{G}) to be the set of all homomorphisms into the multiplicative group of roots of unity in \mathbb{C}. Define a group operation in \check{G} by pointwise multiplication of functions.

  1. Show that this operation on \check{G} makes this set an abelian group. [Hint: show that the trivial homomorphism is the identity and that \chi^{-1}(g) = \chi(g)^{-1}.]
  2. If G is a finite abelian group, prove that \check{G} \cong G. [Hint: write G as \langle x_1 \rangle \times \cdots \langle x_t \rangle and if |x_i| = n_i, define \chi_i to be the homomorphism which sends x_i to e^{2 \pi i/ n_i} and all other x_j to 1. Then show that \chi_i has order n_i and that \check{G} \cong \langle \chi_1 \rangle \times \cdots \langle \chi_t \rangle.]

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 G \rightarrow \check{G}; the isomorphism depends on the choice of generators x_i. This is often stated in the form: a finite abelian group is noncanonically isomorphic to its dual.

  1. Let G be a group and A an abelian group, and denote by H = \mathsf{Hom}(G,A) the set of group homomorphisms G \rightarrow A. We define a product on H pointwise. It is clear that this product is associative. Note that the trivial homomorphism 1 : x \mapsto 1 satisfies (1 \cdot \psi)(x) = 1(x) \psi(x) = \psi(x) and (\psi \cdot 1)(x) = \psi(x) 1(x) = \psi(x), so that 1 is an identity under pointwise multiplication. Now let \psi \in H and define \theta : G \rightarrow A by \theta(x) = \psi(x)^{-1}. Note that \theta(xy) = \psi(xy)^{-1} = \psi(x)^{-1} \psi(y)^{-1} = \theta(x) \theta(y), since A is abelian. So \theta \in H. Moreover, (\theta \cdot \psi)(x) = \psi(x)^{-1} \psi(x) = 1 = 1(x) for all x, so that \theta = \psi^{-1}. Hence H is a group. Finally, note that (\psi \cdot \chi)(x) = \psi(x) \chi(x) = \chi(x) \psi(x) = (\chi \cdot \psi)(x), so that H is abelian.
  2. Define each \chi_i as in the problem statement. Since (by this previous exercise) each \chi_i is a homomorphism, \chi_i \chi_j = \chi_j \chi_i for all 0 \leq i,j \leq t.

    Now let (x_i^{a_i}) \in G. We have \chi_i^{n_i}((x_i^{a_i})) = \prod_{j=1}^t (\chi^{n_i}(x_i))^{a_i} = ((e^{2 \pi i/ n_i})^{n_i})^{a_i} = (e^{\pi i})^{2a_i} = 1^{2a_i} = 1. Thus \chi_i^{n_i} = 1. By this previous exercise, there exists a unique group homomorphism \zeta : G \rightarrow \check{G} such that \zeta(x_i) = \chi_i.

    Note that each homomorphism \chi \in \check{G} is uniquely determined by its values at the x_i (by this previous exercise), and that \chi(x_i) is a n_ith root of unity. Thus \chi(x_i) = \chi_i(x_i)^{a_i} for some a_i (since \chi_i(x_i) is a “primitive n_ith root of unity”, whatever that means.). Consider now \prod x_i^{a_i} \in G; note that \zeta(\prod x_i^{a_i})(x_j) = (\prod \zeta(x_i)^a_i)(x_j) since \zeta is a homomorphism, and that this equals \prod (\zeta(x_i)^{a_i}(x_j)) = \prod \chi_i^{a_i}(x_j) = \chi_j^{a_j}(x_j) = \chi(x_j). By uniqueness, then, and because j is arbitrary, \chi = \zeta(\prod x_i^{a_i}). Thus \zeta is surjective.

    Suppose now that \zeta(\prod x_i^{a_i}) = 1. Then by definition \prod \chi_i^{a_i} = 1. Now let 1 \leq j \leq t; we have 1 = \prod \chi_i^{a_i}(x_j) = e^{2 \pi i a_j/ n_j}. Thus n_j divides a_j, hence a_j \equiv 0 mod n_j. Thus we have \prod x_i^{a_i} = 1, so the kernel of \zeta is trivial and thus \zeta is injective.

    Hence G \cong \check{G}.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: