Constructing a Larger Collection of Functions

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Homework Statement


The groups ##D_3## and ##D_4## are actually collections of functions from the sets ##\{1,2,3\}## and ##\{1,2,3,4\}##, respectively, where those integers represent the vertices of the geometric objects. Is it possible to construct a larger collection of functions from these two sets

Homework Equations

The Attempt at a Solution


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I do not believe this is intended as a rigorous proof, for it is labelled as an "exercise."

At first I thought, "no, it would not be possible, as I actually had to throw away some of the functions/transformations of square in order to get the group ##D_4## because they did not correspond to a rigid transformation of the square." But now I am not so certain this is true; I only threw those other functions/mappings away because they did not correspond to a rigid of the square, but that does not imply that, if they were included in the set ##D_4##, ##D_4## would no longer be a group. If those other functions were included, it most certainly would not be the group of rigid transformation of a square, but that does not mean it is not a group.

Could someone help contrive an approach to solving this problem?
 
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Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
What do you mean by "construct a larger family of functions". "Construct" in what sense? Does this larger family of functions have to satisfy some property?
 
I am sorry. Now that I look back at my original post, I realize that it was very incomplete. Let me try restating the problem a little better.

##D_3## and ##D_4## are two of the dihedral groups, which are actually functions from the set ##\{1,2,3\}## and ##\{1,2,3,4\}## to themselves, respectively, with composition as the group operation. Is it possible to construct a larger collection of functions from these two sets, again with composition as the group operation, and would it still form a larger group?
 
Hint: what is the order of the group of all invertible functions from a finite set to itself?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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