I've noticed, that $S_2 \cong D_1$ and $S_3 \cong D_3$. Is every symmetric group $S_n$ (no including $S_1$) isomorph to the dihedral group $D_{n!/2}$?
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$\begingroup$ related: math.stackexchange.com/q/843447/173147 $\endgroup$– glSCommented May 6 at 7:43
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2 Answers
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No. The symmetric groups $S_n$ for $n\geq 3$ have trivial centre, but the dihedral groups $D_{n!/2}$ all have a centre of order $2$, since $n!/2$ is even.
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James's answer is probably the best. Here's another argument.
Dihedral groups are made of reflections and rotations. That means that there is a morphism $D_n \to \{\pm 1\}$ (the determinant) whose kernel (the rotations) is Abelian.
On the other hand, it is quite easy to show that the only nontrivial morphism $\mathfrak S(n) \to \{\pm 1\}$ is the parity morphism (it suffices to consider the image of the transpositions). So, if $\mathfrak S(n)$ is dihedral, $\mathfrak A(n)$ is Abelian, and that only happens for $n \leq 3$.
