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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.

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    $\begingroup$ The inverse element of $f$ is simply $-f$. $\endgroup$
    – abhi01nat
    Commented Sep 22, 2020 at 12:28

3 Answers 3

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$i)$ 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?

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  • $\begingroup$ Thank you very much for the help! $\endgroup$ Commented Sep 23, 2020 at 7:51
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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 $f$.

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Since there are already detailed explanations, I just add a short remark: Abelian groups are $\mathbb{Z}$-modules. For any $A,B\in \mathrm{obj}(_R \textbf{Mod})$, the set $\mathrm{Hom}_R(A,B)$ is an abelian group (this is easy to see, but one can see the proof in Rotman’s An Introduction to Homological Algebra e2 lemma 2.3). Thus, the set of all homomorphisms from abelian group A to abelian group B is an abelian group.

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    $\begingroup$ The "easy to see" part is essentially the OP's question, right? I am not sure that will help people that much. $\endgroup$ Commented Nov 25, 2023 at 3:01
  • $\begingroup$ I provided a reference for the proof. Since OP's question connects to a larger statement and has been answered in details, so I just want to add a little remark. Hope someone may find it helpful. $\endgroup$
    – Anthony
    Commented Nov 25, 2023 at 6:42

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