# Proving $G^X$ with $(f_1 \cdot f_2)(x) = f_1(x) \circ f_2(x)$ is abelian [closed]

Hello I don't know if i am doing this correct. I find it difficult to prove the statements. I would like some help.

$$G$$ is a group with operation $$\circ$$ and $$X$$ is a non-empty subset.
$$G^X$$ gives the set of all functions $$f: X → G$$.
For $$f_1, f_2 \in G^X$$ we define $$f1 \cdot f_2 \in G^X$$ by $$(f_1 \cdot f_2)(x) = f_1(x) \circ f_2(x) \quad (x \in X)$$.

(a). Show $$G^X$$ is a group with this operation.
To prove $$G^X$$ is a group we have to prove $$G(0)$$ (operation on $$G$$), $$G(1)$$ (associative), $$G(2)$$ (neutral element) and $$G(3)$$ (inverse):

$$G(0)$$: holds as if $$f_1, f_2 \in G^X$$, then the operation $$f_1(x) \circ f_2(x)$$ of the function is also in $$G^X$$

$$G(1)$$: holds. Let $$f_1, f_2, f_3 \in G^X$$. Then
$$(f_1 \cdot (f_2 \cdot f_3))(x) = f_1 \cdot (f_2 \cdot f_3)(x) = f_1 \cdot (f_2(x) \circ f_3(x)) = (f_1 \cdot (f_2(x) \circ f_3(x))(x) = f_1(x) \circ (f_2(x) \circ f_3(x)) = f_1(x) \circ f_2(x) \circ f_3(x) = (f_1(x) \circ f_2(x)) \circ f_3(x) = ((f_1 \cdot f_2)(x))\circ f_3(x) = ((f_1 \cdot f_2)\cdot f_3)(x)$$

$$G(2)$$: If the neutral element is $$1 \in G$$ we get $$(f_1 \cdot 1)(x) = f_1(x) \circ 1 = (1 \cdot f_1)(x) = 1 \circ f_1(x) = f_1(x) \quad \forall f_1 \in G$$

$$G(3)$$ For $$f^{-1}$$ we have $$f^{-1}(x) = f(x)^{-1}$$.
So in the equation we get
$$(f_1 \cdot f_1^{-1})(x) = f_1(x) \circ f_1^{-1} = 1$$ ?

(b). Prove: $$G^X$$ is abelian $$\iff G$$ is abelian.
I don't know. I think we must prove that it is commutative, but I don't know how to continue.

• $\circ$ is not composition here, it is the operation of $G$. Composition of two functions $X\to G$ doesn't make any sense anyway, unless $X=G$. Also, what did you try in the second part? At least one direction should be easy.
– Mark
Commented Feb 19, 2023 at 13:08
• oh yes I see you are right! Commented Feb 19, 2023 at 13:14
• I edited some things Commented Feb 19, 2023 at 13:23
• Note that for the identity you need to clarify the notation. The identity in $G^X$ is the constant function $f(x)=1$, which I believe is what you wanted to write. And again, where are you stuck in the second part? At least the direction that if $G$ is abelian then $G^X$ is abelian is not any more difficult than your proof of associativity.
– Mark
Commented Feb 19, 2023 at 13:36
• Any group has an identity element. You decided to call it $1$, doesn't matter what the operation is. You can call it as you wish. Usually the most common notation is $e$.
– Mark
Commented Feb 19, 2023 at 14:51

Proving that a abelianness of $$G$$ implies abelianness of $$G^X$$ is trivial. For the other direction, since $$X$$ is non-empty, note that for each pair of elements $$a,b \in G$$ there exist constant functions $$f_1, f_2\in G^X$$ mapping the entirety of $$X$$ to $$a,b$$ respectively.
• so i don't have to use the definition that $a \circ b = b \circ a$? Commented Feb 19, 2023 at 13:48
• so i can say if $f_1, f_2 \in G$, then $(f_1 \cdot f_2)(x) = f_1(x) \circ f_2(x) = f_2(x) \circ f_1(x) = (f_2 \cdot f_1)(x)$? Commented Feb 19, 2023 at 13:56
• Yes you can. Assuming abelianness of $G$. Commented Feb 19, 2023 at 14:08
• Ok Thanks! And do you think $G2$ and $G3$ are right, or is the neutral element not equal to $1$? Commented Feb 19, 2023 at 14:15