## 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 $n_i$th 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 $n_i$th 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 $G \cong \check{G}$.