Pontryagin duality for finite groups
Let $\mathbb{Q}$ denote the group of rational numbers (with addition as the binary operation) and let $\mathbb{Z}$ denote the subgroup of integers. The Pontryagin dual of a group $G$ is the group $G^* = \operatorname{Hom}(G,\mathbb{Q/Z})$. (It is most useful when G is abelian).
- Show that $G^*$ is finite if $G$ is a finite group.
- Suppose $G = \mathbb{Z}/n$ for some non-negative integer $n$. Show that $G^*$ is isomorphic to $\mathbb{Z}/n$. Compute $G^*$ (up to isomorphism) for the symmetric group on 3 letters.
for 1 I know if $G$ is finite all its elements are of finite order, and $\mathbb{Q/Z}$ has infinite elements, all of finite order. But I don't know where to go from there or if that's the right idea to begin with. First part of 2 sounds like it follows from similar work that would be done in 1, right now though all I can say is that $G^*$ would be finite. For the second part of 2 I know how the $S_3$ group works but I don't know how to go about finding all the homomorphisms of it to $\mathbb{Q/Z}$, except again, 1 tells me it would be a finite amount. Any help would be appreciated.
Solution 1:
Every element of $\mathbb{Q}/\mathbb{Z}$ can be written uniquely in the form $q+\mathbb{Z}$ for $q\in [0,1)$ (if this isn't obvious, you should take a moment to prove it). So, let $G=\{0,g_1,g_2,\cdots,g_n\}$ be a finite group, and denote $n_i:=\vert g_i \vert$. Then, for all $f\in G^*$, $n_i f(g_i)=0+\mathbb{Z}$. Write $f(g_i)=\frac{a_i}{b_i} + \mathbb{Z}$ with $a_i,b_i\in \mathbb{N}$, $a_i<b_i$, and $gcd(a_i,b_i)=1$. Then, in $\mathbb{Z}$, we have $b_i \vert a_i n_i$ which implies $b_i \vert n_i$. Since there are only finitely many positive integer divisors of $n_i$ and only finitely many corresponding choices for $a_i$, there are a finite number of ways to choose $f$. That takes care of the first question.
For the second, I'll bet you can show that the map sending 1 to $\frac{1}{n}+\mathbb{Z}$ is a generator for $G^*$ by using an argument similar to the one above.
Solution 2:
I got a hint for 1. Yes, $\mathbb Q / \mathbb Z$ has infinitely many elements, but how many elements does it have of order $n$, for given $n$? (Hence these would be the possibilities for elements of $G$ to be mapped to.)
Solution 3:
Here's an alternate solution to Jack's. Suppose first that $G$ is abelian, and decompose it as a direct sum of cyclic groups. Since $\hom(A \oplus B, H) \cong \hom(A, H) \oplus \hom(B, H)$ for any abelian groups $A,B,H$, we are reduced to showing that $2$ is true. Now it suffices to check directly that there are $n$ possible images for the generator of $\mathbb{Z}/n\mathbb{Z}$ (they are $0/n, 1/n, \dots, (n-1)/n$, as Patrick hints to). In fact, this shows that for a finite abelian group $G$, we have $G\simeq G^*$ ('$\simeq$' denotes non-canonical isomorphism).
Now if $G$ is not abelian, notice that by the fact that the abelianization functor is left adjoint to the forgetful functor $\mathbf{AbGrp} \to \mathbf{Grp}$, we have a canonical isomorphism
$$\hom(G_{Ab}, \mathbb{Q}/\mathbb{Z}) \cong \hom(G, \mathbb{Q}/\mathbb{Z})$$
which shows that $\hom(G, \mathbb{Q}/\mathbb{Z})$ is finite for every finite group $G$ (in fact, for every group $G$ whose abelianization is finite).