How many reflection subgroups are in $D_{2n}$?

Question

Given the dihedral group $D_{2n}$ of order $2n$, is there a formula for the number of reflection subgroups of $D_{2n}$?

Answer 1

With a bit of searching, it turns out that that the dihedral group $D_{2n}$ of order $2n$ has exactly $\tau(n) + \sigma(n)$ subgroups, where $\tau(n)$ counts the number of divisors of $n$, and $\sigma(n)$ is the sum of divisors of $n$, evidently (first?) proven by Cavior in 1975, according to this paper by Bivens. You can see a proof of this theorem here, but I'd like to say a bit about why, upon seeing this formula, I believed it had to be correct.

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First, recall that subgroups of $D_{2n}$ are either cyclic, or themselves dihedral.

Next, recall that the cyclic group $C_n$ of order $n$ has a unique subgroup of order $k$ for every $k$ dividing $n$. Hence $D_{2n}$ has $\tau(n)$ cyclic subgroups.

Finally, and this is the best part, we'll think about counting dihedral subgroups of $D_{2n}$. The basic idea is that for a regular $n$-gon, there are exactly $\frac{n}{k}$ regular $k$-gons whose vertices are a subset of the $n$-gon's vertices. For example, thinking about dihedral subgroups of $D_{2 \cdot 12}$, we see that there are exactly $\frac{12}{4} = 3$ vertex-embedded squares in a regular $12$-gon, each giving rise to a subgroup of $D_{2 \cdot 12}$ isomorphic to $D_{2 \cdot 4}$.

Focusing on non-degenerate regular $k$-gons with $k \geqslant 3$, there are at least $$\sum_{k \mid n,\, k \geqslant 3} \frac{n}{k}$$ dihedral subgroups of $D_{2n}$, but this is exactly the sum of all divisors of $n$ that are are at most $\frac{n}{3}$. Of course we have $n$ subgroups of order $2$ generated by reflections, but that still leaves one divisor of $n$, provided $n$ is even (otherwise we're done), $\frac{n}{2}$.

When $n$ is even, for every line $\ell$ of symmetry of the regular $n$-gon there is another line of symmetry $\ell^\perp$ perpendicular to $\ell$. There are exactly $\frac{n}{2}$ such perpendicular pairs. The $\frac{n}{2}$ subgroups $\langle r_\ell,\, r_{\ell^\perp}\rangle$ generated by these reflections across $\ell$ and $\ell^\perp$ are the final dihedral subgroups, bringing our total to $\sigma(n)$.

More is in order to show that these $\sigma(n)$ exhaust all reflection subgroups of $D_{2n}$ (for example: What about stabilizer subgroups of regular polygons whose vertices are midpoints of edges?), but I think that's enough from me.

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Note: Things are a bit nicer if we allow degenerate polygons, $1$- and $2$-gons (which are of course points and line segments). But the symmetry of the argument breaks down a bit, since we would need to think about midpoints of edges of the $n$-gon, instead of just vertices of the $n$-gon.

Answer 2

$D_{2n}$ represents the symmetries of a regular polygon.

If $n$ is odd, then there are $n$ lines of symmetry that intersect one vertex, and the midpoint of the opposite side.

If $n$ is even, there are $\frac n2$ lines through opposite vertices, and $\frac n2$ lines through midpoints of opposite sides.

Any two reflections can act together to form a rotation.

e.g. in $D_8, HV = r^2$ and $\{I, H, r^2, V\}$ forms a subgroup.

There exist pairs of reflections that are capable of generating the entire group.

$D_{2n}$ has a cyclic subgroup of order $n.$ And if $n$ is composite there are subgroups of this cyclic group that correspond to the factors of $n.$

For any reflection line there will be a complementary reflection, that together, will map to a generator of this cyclic subgroup.

Taking $D_{12}$ as our example.

There are the 6 reflection lines.

There are 3 sub-groups of order 4. (That include $r^3$ as a member)

There are 2 sub-groups of order 6 (that include $r^2, r^4$ as members)

There is $D_{12}$ itself.