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Let $X$ be an arbitrary set and let {$G,*$} be a group. I need to show that the set of functions of $X$ onto $G$ where $(f*g)(x)=f(x)*g(x)$ is a group.

I know that to prove that a set is a group, the set must be closed under an associative operation, there must exist an identity for $"*"$, and for every element of $G$, there must exist an inverse with respect to $"*"$.

To prove associativity, let there exist a third element of $G$ called $h(x)$. Then $(f*g)(x)*h(x)=(f(x)*g(x))*h(x)$, and because $f,g,h$ are elements of a group, we can go further and say that $(f*g)(x)*h(x)=(f(x)*g(x))*h(x)=f(x)*(g(x)*h(x))=f(x)*(g*h)(x)$. Thus, $"*"$ is associative on the set.

To show the existence of an identity element, we know that $(f*g)(x)*e=f(x)*g(x)*e=f(x)*g(x)=(f*g)(x)$ where $e$ is the identity element. Hence, $(f*g)(x)*e=(f*g)(x)$, i.e. there exists an identity element for $"*"$ within the set.

Lastly, we show that there exists an inverse element in the set w.r.t. $"*"$. To do this, we note that because $f$ and $g$ are elements of a group, there exists inverse elements $f^{-1}$ and $g^{-1}$ in G for any $f,g$ belonging to G. Thus, $(f*g)(x)=f(x)*g(x)$ implies that: $$(f*g)(x)*g^{-1}(x)*f^{-1}(x)=f(x)*g(x)*g^{-1}(x)*f^{-1}(x)=f(x)*e*f^{-1}(x)=f(x)*f^{-1}(x)=e. $$

And by associativity, $(f*g)(x)*g^{-1}(x)*f^{-1}(x)=(f*g)(x)*(g^{-1}*f^{-1})(x)=e. $ Therefore, for every element of the set, there exists an inverse element w.r.t. $"*"$, and the set is a group. Q.E.D.

Is this a logical proof?

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I do have some criticisms.

  1. Your proof of associativity is fine. A minor, mostly stylistic quibble: you should probably have your chain of equalities start with $((f \star g) \star h)(x) = \ldots$ and end with $\ldots = (f \star (g \star h))(x)$, just so it's obvious that you're showing $(f \star g) \star h = f \star (g \star h)$.

  2. Your proof of identity could use a little work. You have the right idea, but nowhere have you defined an identity element of your proposed group. In particular, you need a function $i : X \to G$ such that $i \star f = f \star i = f$ for all $f$. What is this function $i$? Which values does it take at the various points in $X$? It's not hard, and it's not a trick question, but it's a question that should be answered in your proof.

  3. Similarly, in your proof of inverses, you use $f^{-1}(x)$ and $g^{-1}(x)$, but don't define what they mean. You can't introduce them assuming everyone knows what they are. You are trying to define a group, so you need to construct $f^{-1}$, based on $f$. Which values do they take at the various points in $X$? You need to define them, and show that $f \star f^{-1} = f^{-1} \star f = i$.

Good luck.

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  • $\begingroup$ Thanks for the response. When you say define the inverse, do you mean that I should state "there exists an element of $G$ called $f^{-1}(x)$ such that $f(x)*f^{-1}(x)=e$"? This would be assuming I had already defined $e$. $\endgroup$ – Dion Sep 19 '18 at 14:35
  • $\begingroup$ No, I mean you should specifically define a function $f^{-1}$, so that it multiplies to $f$ to give the identity defined and proven in the previous part of the proof. The identity in the previous part will just be the constant function $i(x) = e$ for all $x \in X$. The inverse of $f$ will be the function that maps $x$ to $(f(x))^{-1}$, where this inverse is the one from $G$. You seem to be confusing $f(x)$, an element of $G$ (whose actual value depends on $x \in X$) with $f$, the function from $X$ to $G$. You need to find an inverse to the function. $\endgroup$ – Theo Bendit Sep 19 '18 at 14:42
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No. You don't distinguish a function $f$ from $X$ into $G$ from an element of $G$. For instance, in the proof of associativity, you assert that “$f$, $g$, $h$ are elements of a group”. If what you are saying is that they are elements of $G$, well, then no, they are not. And if you are saying that they are elements of the group of all functions from $X$ to $G$, then the goal of the exercise is precisely to prove that that's a group.

Associativity: If you have three functions $f,g,h\colon X\longrightarrow G$, then, for each $x\in X$,$$(f(x)*g(x))*h(x)=f(x)*(g(x)*h(x))$$(since $(G,*)$ is a group) and therefore the functions $(f*g)*h$ and $f*(g*h)$ are identical.

Identity element: Take the constant function $x\mapsto e$.

Inverse: Given $f\colon X\longrightarrow G$, consider the map $x\mapsto f(x)^{-1}$.

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  • $\begingroup$ Thank you for the response. Would it be true to say that $f$ is a function, while $f(x)$ is an element of $G$? $\endgroup$ – Dion Sep 19 '18 at 14:23
  • $\begingroup$ @DavidS $f$ is a function from $X$ into $G$, whereas, for each $x$, $f(x)\in G$. $\endgroup$ – José Carlos Santos Sep 19 '18 at 14:24
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Say $G$ is a group, $X$ is a set, $Hom(X,G)$ the set of all mappings from $X$ to $G$.


Definition of a group structure on $Hom(X,G)$

$G$ induces a group structure on $Hom(X,G)$ - in the following way:

Definition (Group Operation)

$(fg):X\rightarrow G$ defined by $(fg)(x):=f(x)g(x)$ for any $f,g\in Hom(X,G)$ and any $x\in X$


Definition (Neutral element)

$1:X\rightarrow G$ defined by $1(x) := 1_G$ for any $x\in X$


Definition (Inverse element)

$g^{-1}:X\rightarrow G$ defined by $g^{-1}(x):=g(x)^{-1}$ for any $g\in Hom(X,G)$ and any $x\in X$


This realy defines a group structure on $Hom(X,G)$

The group axioms for $Hom(X,G)$ are induced by the fact that $G$ is a group - in detail:

(Closure)

$fg \in Hom(X,G)$ for any $f,g\in Hom(X,G)$

since $(fg) (x) = f(x)g(x) \in G$ for any $x\in X$ defines a mapping $X\rightarrow G$ thus $fg\in Hom(X,G)$


(Associativity)

$(fg)h=f(gh)$ for any $f,g,h\in Hom(X,G)$

since $(fg)h (x) = (f(x)g(x))h(x) = f(x)(g(x)h(x)) = f(gh)(x)$ for any $x\in X$


(Inverse Element)

$gg^{-1}=1$ for any $g\in Hom(X,G)$

since $gg^{-1}(x) = g(x)g(x)^{-1}=1_G=1(x)$ for any $x\in X$

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