Can the dihedral groups be seen as subgroups of each other?

Question

I am solving one difficult problem and now I need information on this:

If $m\mid n$, is then possible to "embed" $D_m$ to $D_n$, or otherwise said, does $D_n$ have a subgroup isomorphic to $D_m$?

In my opinion, the answer is yes. This is my "proof" of this (I need you to check if it is correct): As $m\mid n$ there is $k$ such that $mk=n$. Now enumerate the vertices somehow, and take first, then $k+1, 2k+1,...,(m-1)k+1$ vertices. When you join them you shall be getting a regular $m$-tangle, and permutations restricted on it shall be exactly the elements of $D_m$. So, it is a subgroup of $D_n$.

Thanks for any advice (is the statement true, is my proof correct, a correct proof if this one is not correct,...).

Answer 1

Here is an algebraic proof that should help your geometric intuition. The Dihedral group $D_{n}$ has the presentation

$$ \langle x,y \mid x^n = y^2 = (xy)^2 = 1 \rangle $$

Suppose $n = km$. Identify a new element $z = x^{k}$, and consider the subgroup of $D_{n}$ generated by $z$ and $y$. Note that $z^{m} = x^{mk} = x^{n} = 1$. Moreover, $$(zy)^{2} = (x^ky)^2 = x^{k}yx^{k}y = x^{k-1}y^{-1}x^{k-1}y = x^{k-2}yx^{k-2}y = \cdots $$

and we eventually end up with either $y^{-1}y$ or $y^{2}$, both of which are equal to $1$. Then the subgroup generated by $z$ and $y$ has the presentation for $D_{m}$

$$ \langle z,y \mid z^{m} = y^{2} = (zy)^2 = 1 \rangle$$

Technically one would need to check that no other relations hold, but instead one can use an easy order argument and observe that this subgroup has order $2m$, exactly that of $D_{m}$, so no more accidental identifications will occur.

Answer 2

When $m \mid n$, the group $D_n$ definitely has a subgroup isomorphic to $D_m$, and what you've described is precisely why; you have, at least, the heart of the proof.

You can't just enumerate the vertices "somehow," but must be careful to label them cyclically¹, so that vertex $j$ is connected to $j - 1$ and $j + 1$, labels taken modulo $n$. If you do not do this, there are no guarantees that the subset of vertices you have picked out are indeed a regular $m$-gon.

The wording

When you join them you shall be getting a regular $m$-tangle, and permutations restricted on it shall be exactly the elements of $D_m$. So, it is a subgroup of $D_n$.

is a little shaky though. If you mean taking permutations of the full $n$-gon and focusing on what they do to the $m$ vertices you have picked out, you must be careful: You have to focus solely on the permutations that fix the set $\{1 + jk : k = \frac{n}{m},\ j = 0, 1, \ldots, m - 1 \}$. Otherwise, you will have permutations that are not symmetries of your $m$-gon. Depending on the level of rigor, there may still be some nagging questions, like "How exactly do you know this is precisely the set of symmetries of the $m$-gon?" that should be resolved.

It may be cleanest to take the viewpoint of picking vertices of the regular $m$-gon, and focus on its symmetries (which are exactly the permutations in $D_n$ that fix the set $\{1 + jk : j = 0, 1, \ldots, m - 1 \}$ above). Since these also happen to be symmetries of the larger $n$-gon, you have identified a subgroup of $D_n$ that is clearly isomorphic to $D_m$.

If we're thinking primarily in terms of permutations, it would be good to be able to expand on how exactly we produce a symmetry of the $n$-gon from our symmetry of the smaller $m$-gon; that is, explicitly say where vertex $i$ of the $n$-gon gets sent by one of our permutations of vertices of the $m$-gon (if instead we have the full power of "isometries" on our side, then various theorems apply that do the work for us, if we use them carefully). This is the sort of thing that's routinely given as a task "for the reader." It's not exactly hard to do, but it can take some precision.

I think you probably have most of this in mind, but might benefit from trying to make your ideas more explicit.

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¹ Of course you can label the big $n$-gon's vertices however you like; it's your $n$-gon after all :). But unless they are labelled cyclically, you probably won't have the power of modular arithmetic when working with the labels.

Answer 3

A realization of $D_{n}$ that is sometimes useful is that as the following group of permutations of $\mathbb{Z}_{n}$, the set of integers modulo $n$. This is $$ D_{n} = \{ x \mapsto \epsilon x + a : \epsilon = \pm 1, a \in \mathbb{Z}_{n} \}. $$ Let $m \mid n$, so that $n = m k$ for some $k$. Consider the subgroup $H$ of $\mathbb{Z}_{n}$ generated by (the class of) $k$. This is clearly isomorphic to $\mathbb{Z}_{m}$. Now the subgroup of $D_{n}$ given by $$ \{ x \mapsto \epsilon x + a : \epsilon = \pm 1, a \in H \} $$ is a group of permutations of $H$ isomorphic to $D_{m}$.