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It is well known that the symmetries of a triangle, which is the Dihedral Group of order 6, is isomorphic to $S_3$. It is clear that both of these have 6 elements. However, $D_4$, the symmetries of the square, are isomorphic to a subgroup of $S_4$, which has $24$ elements. If some Dihedral Group of order $n$ is equal to $m!$ for some integer $m$, then is that Dihedral Group isomorphic to $S_m$? If so, why?

Shaun
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Ty Jensen
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    No. $D_{6}$ and $S_{3}$ are the only isomorphic pair of dihedral/symmetric groups. I'm guessing this might be a homework problem, so I will encourage you to think about why this is so instead of telling you. – xxxxxxxxx Mar 20 '20 at 23:46
  • @MorganRodgers just for the record, this isn’t a homework problem, but I will think more about this as you are encouraging! Thanks. – Ty Jensen Mar 20 '20 at 23:48
  • @MorganRodgers Does the solution have to do with the fact that Dihedral Groups with order $m!$ with $m>3$ are not bijective? I can certainly picture this in my head to a degree, but I can't think of how to describe it rigorously. – Ty Jensen Mar 20 '20 at 23:59
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    I'm not sure what it would mean for a group to be bijective. Do you mean doubly transitive? The easiest way to see they are not isomorphic is to consider their largest cyclic subgroups. – xxxxxxxxx Mar 21 '20 at 00:20

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$S_n$, unlike the dihedral group, has a unique subgroup of index two, $A_n$, that is never abelian ($n\gt3$).

The dihedral group, $D_m$, contains a subgroup of index two that is cyclic. A group of rotations.

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All dihedral groups are solvable.

For $n \ge 5$, the symmetric group of degree $n$ is non solvable.

For $n=4$, note that every element of $S_4$ has order at most $4$, while $D_{12}$ has an element of order $12$, whence $S_4 \ncong D_{12}$.

Therefore, $D_m \cong S_n \iff n=m=3$.