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Problem: Assume $S$ is a nonempty set and $G$ is a group. Let $G^S$ denote the set of all mappings from $S$ to $G$. Find an operation on $G^S$ that will yield a group.

Update (full attempted proof--is it correct?): Let $*$ be the operation on the given group $G$ and let $\bigstar$ be the operation on $G^S$. For all $\alpha,\beta\in G^S$, define $(\alpha\,\bigstar\,\beta)(x)=\alpha(x)*\beta(x)$. Now we must prove $G^S$ is a group.

  • Closure: This is inherited from $G$.
  • Associativity: This is also inherited from $G$.
  • Existence of an identity element: This is also inherited from $G$; more explicitly, $e_G(x)=1=e_{G^S}(x)$ because $(\alpha\,\bigstar\,e_G)(x)=\alpha(x)*e_G(x)=\alpha(x)$ and, in the other direction, $(e_G\,\bigstar\,\alpha)(x)=e_G(x)*\alpha(x)=\alpha(x)$.
  • Existence of inverse elements: If $\eta(x)\in G$, then $\eta^{-1}(x)\in G$ also. Thus, for $\alpha\in G^S$, we have $(\alpha\,\bigstar\,\alpha^{-1})(x)=\alpha(x)*\alpha^{-1}(x)=e_G(x)=e_{G^S}(x)$ and, in the other direction, $(\alpha^{-1}\,\bigstar\,\alpha)(x)=\alpha^{-1}(x)*\alpha(x)=e_G(x)=e_{G^S}(x)$.

This concludes the proof that the operation $\bigstar$ on $G^S$ yields a group. $\blacksquare$


$\color{red}{\mathbf{\text{Question:}}}$ Is this above proof correct? Also, by the way the operation $\bigstar$ works, it seems like $G^S$ would be a subgroup of $G$ or am I wrong?


Original work shown for problem:

Book solution: For $\alpha,\beta\in G^S$, define $(\alpha\beta)(x)=\alpha(x)\beta(x)$ for each $x\in S$.

My questions: I thought about this problem for a good while but got nowhere and finally decided to look at the answer, but it does not make a ton of sense to me. How does the given operation yield a group? And what really is the operation being considered? Is it composition? That is, $(\alpha\beta)(x)\equiv(\alpha\circ\beta)(x)$? If it is composition, then I know all compositions are associative, and thus I do not need to prove that as part of proving that $G^S$ is a group. But what about closure? Existence of an identity element? Existence of inverse elements?

For the existence of an identity element, I thought about $e_G(x)=1$ and having $(\eta e_G)(x)=\eta(x)e_G(x)=\eta(x)$ and $(e_G\eta)(x)=e_G(x)\eta(x)=\eta(x)$, but this doesn't really seem right or is it?

Lastly, the problem gives the condition that $S\neq\varnothing$ (why is that important?), and no operation is specified concerning $G$--does that matter? I guess not, but it seems interesting to me that no operation is specified for $G$ when we are trying to find an operation for $G^S$ that involves $G$.

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  • $\begingroup$ The proof seems fundamentally fine. I'm not sure what the distinction is between the notation $e_G(x)$ and $e_G^S(x)$. You could write out associativity more explicitly, although I tend not to do so unless it's not obvious. $\endgroup$ – Rolf Hoyer Apr 9 '15 at 17:22
  • $\begingroup$ @RolfHoyer I just meant that $e_G(x)$ is the identity element for $G$ while $e_G^S(x)$ is the identity element for $G^S$. If $G^S$ is a subgroup of $G$ (it looks like it is, but is it?), then I know the distinctions do not matter because the identity and inverses are the same. $\endgroup$ – fancynancy Apr 9 '15 at 17:32
  • $\begingroup$ @RolfHoyer OK--I just fixed up the question with the modified notation ($e_{G^S}(x)$ instead of $e_G^S(x)$). Look all good now? Your answer definitely helped me to start thinking along the right lines. $\endgroup$ – fancynancy Apr 9 '15 at 17:38
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It definitely looks much better!

I'm a little confused about the distinction between $e_G$ and $e_{G^S}$; they both seem like the identity of $G^S$, and you seem to be using $1$ for the identity of $G$. Maybe make it more explicit that you're defining $e_{G^S}: S \to G$ so that $e_{G^S}(x) = 1_G$; it's the map sending everything in $S$ to the identity of $G$. So my only suggestion would be to clean up the notation (how exactly are you writing the identity of $G$? Of $G^S$?) and make it clear exactly what the identity of $G^S$ does.

For the inverses, essentially the same thing: Given $\alpha: S \to G$, you just start using $\alpha^{-1}$ without explicitly stating that, if $\alpha(x) = g$, then $\alpha^{-1}(x) = (\alpha(x))^{-1} = g^{-1}$. That is, make it clear that we invert $\alpha:S \to G$ pointwise, by inverting its output (using the inverse in $G$).

For your other question, very rarely will it be the case that $G^S \cong G$. I think it's more likely that $G^S$ is isomorphic to a direct product of several copies of $G$ (how many copies?).

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  • $\begingroup$ Hmm okay. So I should write $e_G(x)=1$ and $e_{G^S}(x)=1_G$ to more clearly indicate that $1$ is the identity for $G$ while $1_G$ is the identity for $G^S$ right? You're right--my notation is a bit mangled. Okay. I get the identity part now. I'm still a little fuzzy on the inverse part though. Are you saying I'm technically correct with what I've written but it would be clearer to actually assign an element to $\alpha(x)$ to clearly indicate that invertibility is pointwise? $\endgroup$ – fancynancy Apr 9 '15 at 18:10
  • $\begingroup$ What I mean about identities is that there should really only be two in sight: That of $G$, let's call it $1_G$, and that of $G^S$, let's call it $1_{G^S}$. So, I'm not sure why there are two functions whose notation suggest that they're both identities of $G^S$; that they're both functions $S \to G$. $\endgroup$ – pjs36 Apr 9 '15 at 18:12
  • $\begingroup$ And for the inverses, yes, it's correct, provided you explicitly state how $\alpha^{-1}: S \to G$ (or $\eta^{-1}$, for that matter) is defined. $\endgroup$ – pjs36 Apr 9 '15 at 18:14
  • $\begingroup$ Okay so I should have (for identities, that is), in one direction, $(\alpha\,\bigstar\,e_G)(x)=\alpha(x)*e_G(x)=\alpha(x)*1_G=\alpha(x)$. The same would happen in the other direction. So the identity exists and, explicitly, we have $e_{G^S}=e_G=1$. Correct? $\endgroup$ – fancynancy Apr 9 '15 at 18:16
  • $\begingroup$ I do see what you're saying now about the confusion between writing $e_G(x)$ and $e_{G^S}(x)$ because they are both mappings from $S$ to $G$; thus, that notation is rather clumsy. $\endgroup$ – fancynancy Apr 9 '15 at 18:19
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The group operation is not defined by composition, unless you mean the composition $S \to S\times S \to G\times G \to G$, where the first map is the diagonal $s \to (s,s)$, the second map is the product of the two given maps $\alpha, \beta$, and the third map is the group operation on $G$. When the author writes $(\alpha\beta)(x) = \alpha(x)\beta(x)$, the latter product is defined using the operation of $G$, just to clarify.

This definition is using the operation of $G$ pointwise, just like how addition and multiplication are defined for real-valued functions, ie $(f+g)(x) = f(x) + g(x)$. The identity element is the constant function $e_G(x) = 1$ for every $x$, and you correctly give the proof that this works. The inverse of a function $\alpha: S \to G$ is given by taking inverses pointwise, namely $\alpha^{-1}(x) = (\alpha(x))^{-1}$. Associativity follows from the associativity of $G$.

As regards to the empty set, if $S = \emptyset$ then there is a unique empty function $S \to G$, and I guess the authors would rather rule out this case than give the trivial product on this set.

Hopefully this makes things more clear.

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  • $\begingroup$ I just updated my question with a fully attempted proof in light of your helpful answer. Can you let me know if you think it's correct? Thanks for your answer--I think I am at least on the right track now. $\endgroup$ – fancynancy Apr 9 '15 at 14:23

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