Set of Homomorphisms between two abelian groups is a group

Question

Let $(G,+)$,$(G^\prime,+)$ two abelian groups and $H, H^\prime$ two proper subgroups of $G$ and $G^\prime$ respectively. Let $\text{Hom}(G,G^\prime) = \{f:G \rightarrow G^\prime : f\text{ homomorphism}\}$. Prove that $\text{Hom}(G,G^\prime)$ is a group endowed with the operation definded by:

$$(f+g)(x) = f(x)+g(x), \forall x\in G$$

I thought it would be easily proven by showing that the group axioms hold. But trying this:

$i)$ Closure: Let $f,g\in \text{Hom}(G,G^\prime)$. Then: $$(f+g)(x) = f(x) + g(x)$$ But I don't know how to show that $f(x)+g(x)$ could be again in $\text{Hom}(G,G^\prime)$. Do I have to somehow use the fact that $G^\prime$ is commutative?

$ii)$ Associativity: Let $f,g,h\in \text{Hom}(G,G^\prime)$. Then: $$(f+g)(x)+h(x) = f(x)+g(x)+h(x)=f(x)+(g+h)(x)$$

$iii)$ Identity element: Let $f\in \text{Hom}(G,G^\prime)$. $$(f+0)(x) = f(x) + 0(x) = 0(x)+f(x) = f(x)$$

$iv)$ Inverse element: Let $f \in \text{Hom}(G,G^\prime)$. Then: $$0 = \dots?$$

I'm kind of stuck and don't know how to approach the last. Any help or advise would be appreciated. Thanks.

Answer 1

$i)$ Closure: Let $f,g\in Hom(G,G^\prime)$. Then: $$(f+g)(x) = f(x) + g(x)$$ But I don't know how to show that $f(x)+g(x)$ could be again in $Hom(G,G^\prime)$. Do I have to somehow use the fact that $G^\prime$ is commutative?

Indeed. So we have two group homomorphisms $f,g\in Hom(G,G^\prime)$ and we already know how to add them. So lets check that the result is a homomorphism:

$$(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)=$$ $$=f(x)+g(x)+f(y)+g(y)=(f+g)(x)+(f+g)(y)$$

Do you see where we've applied commutativity?

$ii)$ Associativity: Let $f,g,h\in Hom(G,G^\prime)$. Then: $$(f+g)(x)+h(x) = f(x)+g(x)+h(x)=f(x)+(g+h)(x)$$

That looks ok. To be formally correct you have to write the initial and final conditions, i.e. you should start with $((f+g)+h)(x)$ and end with $(f+(g+h))(x)$.

$iii)$ Identity element: Let $f\in Hom(G,G^\prime)$. $$(f+0)(x) = f(x) + 0(x) = 0(x)+f(x) = f(x)$$

Again, that looks fine. You just missed the definition: $0(x):=0_{G^\prime}$.

$iv)$ Inverse element: Let $f \in Hom(G,G^\prime)$. Then: $0 = ...?$

How about $g(x):=-f(x)$? Can you complete the proof?

Answer 2

You don't need to show that $f(x)+g(x)$ is an element in $Hom(G,G')$, this is of course not true. You need to show that the function $f+g$ is a function in $Hom(G,G')$, i.e you have to show that $f+g$ is a homomorphism. And indeed:

$(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)=(f(x)+g(x))+(f(y)+g(y))$

$=(f+g)(x)+(f+g)(y)$

We used commutativity here to change the order of summation. So $f+g$ is indeed a group homomorphism.

The identity element is indeed the function which maps every element of $G$ to $0$. As for the inverse: given $f\in Hom(G,G')$ define a function $g:G\to G'$ by $g(x)=-f(x)$. This is a homomorphism because:

$g(x+y)=-f(x+y)=-(f(x)+f(y))=-f(x)-f(y)=g(x)+g(y)$

And hence $g\in Hom(G,G')$. Also, for every $x\in G$ we have:

$(f+g)(x)=f(x)+g(x)=f(x)-f(x)=0=0(x)$.

And similarly $(g+f)(x)=0(x)$. So $g$ is the inverse of $f$.