Exercises on character group $G^*$ of a group $G$

Question

Let $S^1=\{z \in \mathbb{C} : |z|=1\}$, which is a group under multiplication of functions. For a group $G$ define $G^*=Hom(G,S^1)$, the character group of $G$. Prove that $G^*$ is indeed a group under multiplication of functions. Prove:

$a) (A\oplus B)^* \cong A^* \oplus B^*.$

$b)$ If $G$ is a finite abelian group then $G \cong G^*$.

$c)$ Let $G$ be a finite abelian group and $H$ a subgroup of $G$. Show that there is a subgroup $N$ of $G$ such that $G/N \cong H$. Similarily, if $H$ is isomorphic to a quotient group $G$ then $H$ is isomorphic to a subgroup of $G$.

I need help with c), I am not sure if a),b) are correct.

Solution: First we show $G^*$ is a group under multiplication.

Identity: $\mathbb{1}:= \mathbb{1}(a)=1 \in G^*$ is the identity element in $G^*$ as if $f \in G^*$, $(f*\mathbb{1})(a)=f(a)*1=f(a)=1*f(a)=(\mathbb{1}*f)(a)$.

Associativity: let $f,g,h \in G^*$, $((f*g)*h)(a)= (f(a)*g(a))*h(a)=f(a)*(g(a)*h(a))=(f*(g*h))(a)$.

Inverse: Let $f \in G^*$, then let $f^{-1}(a):=f(a)^{-1}$, we show $f^{-1}$ is a homomorpism: $f^{-1}(0)=f(0)^{-1}=1^{-1}=1$, $f^{-1}(a+b)=f(a+b)^{-1}=(f(a)f(b))^{-1}=f(a)^{-1}f(b)^{-1}=f^{-1}(a)f^{-1}(b)$.

Hence, $G^{*}$ is a group under multiplication.

a) Consider the map $\phi: (A\oplus B)^* \rightarrow A^* \oplus B^*$ by $\phi(f)=(f|_{A},f|_B)$, it is an isomorphism.

b) Let $G$ be a finite abelian group, then $G \cong \oplus_{1}^{n}\mathbb{Z}_p$, by $2)$, $G^* \cong \oplus_{1}^{n}\mathbb{Z}_p^* \cong \oplus_1^n\mathbb{Z}_p$.

Answer 1

I don't know how much you know about category theory and homological algebra, but I'm going to prove the first part of c) showing that $Hom(-,S^1)$ is a contravariant right-exact functor from the category of finite abelian groups to itself.

That means that given the exact sequence $0\to H\to G\to G/H$, it will give us an exact sequence $Hom(G/H,S^1)\to Hom(G,S^1)\to Hom(H,S^1)\to 0$. And since $Hom(-,S^1)$ is the ''identity'' on objects, i.e., $Hom(G,S^1)\cong G$, we will get an exact sequence $G/H\to G\to H\to 0$ (it actually gives an isomorphic exact sequence, but exactness of one follows from exactness of the other one). Since you know the definition of exact sequence, I'm assuming that you know that this mean that $H\cong G/\mathrm{Im}(G/H)$ using the first isomorphism theorem, so $N= \mathrm{Im}(G/H)$.

Given a homomorphism $h:H\to G$, there is a natural homomorphism $h^* =Hom(h,S^1):G^*\to H^*$ given by $h^*(g)=g\circ h$. With this in mind, I'll let you check that $Hom(-,S^1)$ is indeed a contravariant functor using the definitions.

Now, we already have that $Hom(-,S^1)$ applied to $0\to H\to G\to G/H $ gives us a (not yet exact) sequence $G/H\to G\to H\to 0$, since it reverses arrows and respects compositions in the contravariant sense. Then, let's prove that $Hom(-,S^1)$ transforms the injection $H\to G$ into a surjection $G\to H$.

To show that $h^*$ is surjective I take two homomorphisms $j: F\to H$ and $k:F\to H$ such that $j^*h^*=k^*h^*$. Let's apply this to $z\in \operatorname{Hom}(F,S^1)$. We get $j^*h^*(z)=k^*h^*(z)\Rightarrow z\circ h\circ j=z\circ h\circ k$. Now take and injective $z$, which exists since the image of $(1,\dots, 1)$ under the isomorphism $F\cong Hom(F,S^1)$ is injective. To see this, remember that an element of $Hom(F,S^1)$ is determined by the $p_i$-th root of unity that is the image of $1\in\mathbb{Z}_{p_i}$. Since $F=\oplus_{i=1}^n\mathbb{Z}_{p_i}$ being $p_i\neq p_j$ for $i\neq j$, the only $p_i$-th root that is $p_j$-root is $1$. Then, by injectivity of $z$, $h\circ j=h\circ k$, and by injectivity of $h$, $j=k$, hence $j^*=k^*$ and therefore $h^*$ is surjective.

This shows that $G\to H\to 0$ is exact. Finally, we need to show that $\ker{p^*}=\mathrm{Im}{h^*}$, where $p:G\to G/H$ is the quotient map. This follows from the more general fact that $Hom(-,X)$ is left-exact in many contexts. It is not a hard proof so you could try to do it, but you can also look it up here in the case of $R$-modules (the proof for finite abelian groups is similar).

There are more general and abstract ways to prove this fact, for example the one in this answer, but since you don't know a lot about category theory I tried to use less of that machinery.

Answer 2

Let $i \colon H \to G$ be the inclusion homomorphism. We can define the homomorphism $i^* : G^* \to H^*$ as $i^*(f) = f \circ i$. @Javi proves that $i^*$ is surjective in the fifth paragraph of his answer. Apply the first isomorphism theorem to get $G^*/\ker i^* \cong H^*$. From (b), you can remove the stars from $G^*$ and $H^*$. You can also deduce from (b) that there is a subgroup $N < G$ with $N \cong N^* \cong \ker i^*$.