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In M. Artin's book Algebra he wrote: enter image description here

But I think this visualisation $D_1$ and $D_2$ is inconsistent and confusing, because I guess $D_n$ could be associated with a subgroup of $S(\{1, \ldots, n\})$, i.e. permutations of the $n$ vertices, but in the case of the $1$-gon there is just one permutation, the identity, and in the case of the $2$-gon there are just two (identity and one transposition). Furthermore by a reflection of the $2$-gon in the horizontal axis it gets projected onto itself (every vertex stays where it is), so this reflection is actually the identity, and not a separate element $r \ne 1$. Or does I miss something? Could it be consistent to look at $D_1$ and $D_2$ in this way?

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up vote 4 down vote accepted

What Artin is saying is excactly that

because I guess $D_n$ could be associated with a subgroup of $S(\{1,\ldots,n\})$, i.e. permutations of the $n$ vertices

is not true for $D_1$ and $D_2$ (at least with the definition he uses for that notation).

By the symmetry of the $n$-gon you should, in this case, think of the group of all the isometries of the plane that fix the outline of the $n$-gon. For $n=1$ there are two such isometries, namely the identity and the reflection through the midline.

It's a general fact about isometries of the plane that knowing what they do to three non-collinear points is enough to reconstruct the entire isometry. In particular for $n\ge 3$ it just happens that knowing what happens to each vertex of an $n$-gon is enough to reconstruct an entire isometry, so in that particular case one can represent elements of $D_n$ by permutations from $S_n$. But that's just a practical property that holds in the case $n\ge 3$, not part of the definition of $D_n$.

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Then you're violating the consistent pattern that $D_n$ is a group of order $2n$. Since the dihedral group is yiven predictably in terms of generators and relations, this should be inviolable. If you like, knowing what happens just to the vertices won't determine the symmetry for $n=1$ and $2$.

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1 and 2 are the only positive integers $n$ for which $2n \nmid n!$, which is why $D_2$ and $D_4$ are not subgroups of $S_1$ and $S_2$.

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why downvote?!! – S. Snape Jul 22 '14 at 14:33

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