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Obviously one can let the dihedral group of order 8 act on a square. How do I define this action mathematically correct? Since the dihedral group leaves the square invariant and just permutes the corners, must I take the set $M=\{(a,b,c,d)\mid 1\leq a,b,c,d\leq4\}$? And then i.e. for the rotation $r\star (1,2,3,4)=(4,1,2,3)$?

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    $\begingroup$ @anon Your comment of "why would you consider such a set" sounds needlessly dismissive - the OPs idea is actually quite a natural way to think of the action. The element $(4, 1, 2, 3)$ corresponds to the permutation $(4, 1)(1, 2)(2, 3)(3, 4)$ naturally; the OPs interpretation corresponds to the "standard" idea of a Cartesian product, while the permutation interpretation corresponds to thinking of a Cartesian product using an index set ($4\mapsto 1, 1\mapsto 2, 2\mapsto 3, 3\mapsto 4$). $\endgroup$ – user1729 Aug 29 '17 at 8:10
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    $\begingroup$ @anon That is not a permutation. $(a, b, c, d)$ corresponds to a permutation if and only if the numbers $a$, $b$, $c$ and $d$ are pairwise distinct. $\endgroup$ – user1729 Aug 29 '17 at 15:30
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    $\begingroup$ @anon Okay, I think I see your point. Yes, $(1, 1, 1, 1)$ is part of the OPs set, but one can ignore it and assume it is fixed. This is what I am doing (and, rightly or wrongly, I presumed that this is what the OP is doing). However, if you nitpick about this then you would have to nitpick about this group acting on the set with $5$ elements! I am merely trying to say that the OPs interpretation is natural, and not weird as you are suggesting. $\endgroup$ – user1729 Aug 30 '17 at 8:56
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    $\begingroup$ @anon I disagree - it is mathematically correct! It is not $X^n$ they are considering, but rather $X^4$. This $4$ is important. Their thought is entirely analogous to the idea that the Caresian product $X^4$ is the set of functions $\{f: \{1, 2, 3, 4\}\rightarrow X\}$. $\endgroup$ – user1729 Aug 30 '17 at 14:36
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    $\begingroup$ @anon Yes, sure, but the OP has encoded a square in a very specific, and quite natural, way . $\endgroup$ – user1729 Aug 30 '17 at 14:59
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A group action is merely a group homomorphism from a group $G$ to a subgroup of $\text{Sym}(X) \cong S_{|X|}$. It appears you want a monomorphism; that is, essentially "another way of describing $G$".

Since $|D_4| = 8$, to obtain such a monomorphism, we require $|X| > 3$. So expressing $D_4$ is a permutation of "$4$ somethings" is going to be optimal. What (symmetric) aspects of a square come in fours?

To understand more fully what I am getting at, note that the square has two diagonals, and that $D_4$ can act on these diagonals (how?). If we see $D_4$ as a subgroup of $S_4$, what are your conclusions regarding $D_4 \cap A_4$?

One of the comments suggest you might see $D_4$ as a subgroup of $\text{GL}_2(\Bbb R)$. To get you started how does:

$\rho = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$

act on the set $X = \{(1,0), (0,1), (-1,0), (0,-1)\}$?

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