1
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ $\circ$ 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. $\endgroup$
    – Mark
    Commented Feb 19, 2023 at 13:08
  • $\begingroup$ oh yes I see you are right! $\endgroup$
    – Willem
    Commented Feb 19, 2023 at 13:14
  • $\begingroup$ I edited some things $\endgroup$
    – Willem
    Commented Feb 19, 2023 at 13:23
  • $\begingroup$ 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. $\endgroup$
    – Mark
    Commented Feb 19, 2023 at 13:36
  • 1
    $\begingroup$ 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 $e$. $\endgroup$
    – Mark
    Commented Feb 19, 2023 at 14:51

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ so i don't have to use the definition that $a \circ b = b \circ a$? $\endgroup$
    – Willem
    Commented Feb 19, 2023 at 13:48
  • 1
    $\begingroup$ What do you mean? You do have to use it in proving the trivial direction… $\endgroup$
    – Vivaan Daga
    Commented Feb 19, 2023 at 13:49
  • $\begingroup$ so i can say if $f_1, f_2 \in G$, then $(f_1 \cdot f_2)(x) = f_1(x) \circ f_2(x) = f_2(x) \circ f_1(x) = (f_2 \cdot f_1)(x)$? $\endgroup$
    – Willem
    Commented Feb 19, 2023 at 13:56
  • $\begingroup$ Yes you can. Assuming abelianness of $G$. $\endgroup$
    – Vivaan Daga
    Commented Feb 19, 2023 at 14:08
  • $\begingroup$ Ok Thanks! And do you think $G2$ and $G3$ are right, or is the neutral element not equal to $1$? $\endgroup$
    – Willem
    Commented Feb 19, 2023 at 14:15

Not the answer you're looking for? Browse other questions tagged .