What is the symmetry point group of a regular n-gon, where n is even with the opposite vertices identified? Is it $D_n$?

Question

If we have a regular 2n-gon, what is its symmetry group if we identify the opposite vertices (the ones that are on the same line). Do we get the group that is isomorphous to the group of the regular n-gon?

Answer 1

It is quite obvious to see, that the group you search is $\frac{D_{2n}}{H_{2n}}$, where $H_{2n}$ is the group of all symmetries of the $2n$-gon that send each vertices either to itself, or to a diametrically opposite one. Let's find $H_{2n}$.

If a symmetry preserves three vertices, then it is known to be trivial. If it preserves two non-opposite vertices, it is also trivial. And if it preserves two opposite vertices it is either trivial or a "mirror" symmetry in respect to the diagonal connecting them. It satisfies our conditions only if $n = 2$ and your $2n$-gon is a square. If the symmetry preserves one vertice, it maps its neighbour vertices to its neighbour vertices, which results init being a "mirror" symmetry in respect to a diagonal. We have already examined that case. Now, if it preserves nothing, then to satisfy our condition it has to be the central symmetry.

Now we have classified the elements of $H_{2n}$ and now we see that there are two different cases:

1) $n = 2$

$H_4 \cong C_2 \times C_2$ and so your group is $C_2$

2) $n > 2$

Then $H_{2n}$ is the cyclic group generated by central symmetry (which is actually a rotation on $\pi$), so $\frac{D_{2n}}{H_{2n}} \cong \frac{\langle a \rangle_{2n} \rtimes \langle b \rangle_2}{\langle a^n \rangle} \cong D_n$

So, your group is $C_2$ in case $n = 2$, and $D_n$ otherwise.