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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:

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

$b)$ 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 $phi: (Aoplus B)^* rightarrow A^* oplus B^*$ 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 $2)$, $G^* cong oplus_{1}^{n}mathbb{Z}_p^* cong oplus_1^nmathbb{Z}_p$.

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 $0to Hto Gto G/H$, 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 $G/Hto Gto Hto 0$ (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 $Hcong G/mathrm{Im}(G/H)$ using the first isomorphism theorem, so $N= mathrm{Im}(G/H)$.

Given a homomorphism $h:Hto G$, there is a natural homomorphism $h^* =Hom(h,S^1):G^*to H^*$ given by $h^*(g)=gcirc h$. 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 $0to Hto Gto G/H $ gives us a (not yet exact) sequence $G/Hto Gto Hto 0$, since it reverses arrows and respects compositions in the contravariant sense. Then, let's prove that $Hom(-,S^1)$ transforms the injection $Hto G$ into a surjection $Gto H$.

To show that $h^*$ is surjective I take two homomorphisms $j: Fto H$ and $k:Fto H$ such that $j^*h^*=k^*h^*$. Let's apply this to $zin operatorname{Hom}(F,S^1)$. We get $j^*h^*(z)=k^*h^*(z)Rightarrow zcirc hcirc j=zcirc hcirc k$. Now take and injective $z$, which exists since the image of $(1,dots, 1)$ under the isomorphism $Fcong Hom(F,S^1)$ 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 $1inmathbb{Z}_{p_i}$.
Since $F=oplus_{i=1}^nmathbb{Z}_{p_i}$ being $p_ineq p_j$ for $ineq j$, the only $p_i$-th root that is $p_j$-root is $1$. Then, by injectivity of $z$, $hcirc j=hcirc k$, and by injectivity of $h$, $j=k$, hence $j^*=k^*$ and therefore $h^*$ is surjective.

This shows that $Gto Hto 0$ is exact. Finally, we need to show that $ker{p^*}=mathrm{Im}{h^*}$, where $p:Gto G/H$ 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.

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 $N cong N^* cong ker i^*$.