Embedding of smaller order dihedral group, $D_a$, inside bigger: $D_b,$ s.t. $a\mid b.$

Question

Request vetting, and a question at the end. Let $a=2, b=10.$

There can be four elements in $D_4$ hence three non-trivial elements in any such subgroup (corresponding to the embedding) in $D_{10}.$

Reflection elements exist in pairs, as composition yields rotation. But, leaving three elements means rotation element is there and is derived by composition of the two reflections.

Also, need for closure is satisfied by:
$sr^a.r^{b-a}= sr^b, sr^b.r^{b-a}\equiv sr^b.r^{a-b}=sr^a, sr^a.sr^b = r^{b-a}.$

There cannot be possibilities of embedding $D_2$ inside any $D_n, 2\nmid n$ (as order $4$ subgroup cannot embed inside an odd order group, by Lagrange theorem).

Isn't it wonderful that the algebraic equations describing dihedral object $D_n$ can describe all possible embeddings too, also intuitively?

Request some good text, old or new, that describes embedding of smaller dihedral objects inside bigger ones. Or, some link.

But, still want to know if any embedding of $D_2$, inside $D_n$, is possible on the edge joining two opposite vertices?

Answer 1

I don't have any text to recommend, I just have my own geometric intution. I find that geometry helps me to understand $D_n$, and its embedded $D_2$'s.

Here's a geometric classification of $D_2$ subgroups of $D_n$, equating $D_n$ with the symmetry group of a regular $n$-gon.

And as you say, it is necessary that $n$ be even (because $D_2$ has order $4$ and if $n$ is odd then $4 \nmid 2n$).

Case 1: $4 \nmid n$. Start with any diameter $L_1$ of the $n$-gon that joins opposite vertices. Next let $L_2$ be the perpendicular diameter that bisects an opposite pair of edges. The reflections across $L_1$ and $L_2$ generate a $D_2$ subgroup of $D_n$. All $D_2$ subgroups of $D_n$ arise in this manner (in Case 1).

Case 2: $4 \mid n$. There are two types of $D_2$ subgroups of $D_n$.

1. Type V: Choose any perpendicular pair of diameters $L_1,L_2$ each of which joins opposite Vertices.
2. Type E: Choose any perpendicular pair of diameters $L_1,L_2$ each of which bisects an opposite pair of Edges.

For each of these two types, reflections across $L_1$ and $L_2$ generate a $D_2$ subgroup of $D_n$. All $D_2$ subgroups of $D_n$ arise in this manner (in Case 2).

So, for example, in $D_4$, which is the symmetry group of a square, there are two $D_2$ subgroups: one of Type V; and one of Type E.

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To summarize (assuming I understand your terminology), in addition to considering diameters that join an opposite pair of vertices, in order to get a complete description of $D_2$ subgroups of $D_n$ one must also consider diameters that bisect an opposite pair of edges.

So your question can be answered with a "Yes": a $D_2$ subgroup of $D_n$ can be described starting with any diameter of the $n$-gon that joins opposite vertices.

But on the other hand, it is not true that every $D_2$ subgroup of $D_n$ can be described starting in that manner, because when $4 \mid n$ the Type E subgroups can only be described using diameters that bisect opposite pairs of edges.