# Regarding $D_{2m}$ as a subgroup of $S_n$ [closed]

I already know that the Dihedral group $$D_{2n}$$ can be regarded as a subgroup of $$S_{n}$$ by regarding a symmetry of the $$n$$-gon as a permutation of $$n$$ verticies. In the same manner, can I regard the group $$D_{2m}$$ where $$m as a subgroup of $$S_n$$? (permutation of $$m$$ verticies and fix other $$n-m$$ elements)

• Yes. Note that some people label the symmetries of the $n$-gon as $D_n$, and others label it $D_{2n}$. Aug 16 at 6:46
• "can I regard $D_{2m}$ as a subgroup of $S_n$" is quite sloppy. The correct formulation is "is $D_{2m}$ isomorphic to $S_n$". And actually there can be several non-equivalent such embeddings.
– YCor
Aug 16 at 18:26
• Conversely it can easily be shown that $D_{2n}$ is not isomorphic to any transitive subgroup of $S_m$ if $m<n$. (Exercise: show that $D_{42}$ is isomorphic to a subgroup of $S_{20}$ although $20<42/2$.)
– YCor
Aug 16 at 18:30
• If $D_{42}$ is the dihedral group on $21 = 7\cdot 3$ points then it's isomorphic to a subgroup of $S_{10}$ ($10=7+3$), but not $S_9$. Aug 16 at 18:57
• In case people are interested in Robert's comment: it turns out for $m = \prod p_i^{k_i}$, the least $n$ with $D_{2m} \hookrightarrow S_n$ is $\sum p_i^{k_i}$. You can find a proof on my blog, which also links to the mse question where I first saw this proven. Aug 16 at 19:51

You're completely right! In the interest of making this answer slightly less trivial, I'll say some more words about your idea, as well as some more advanced mathematics that it's pointing towards.

First, we should notice that a symmetry of an $$m$$-gon is completely determined by where the vertices get sent. After all, if we know where the endpoints of an edge go, then there's only one place to send the edge.

This means $$D_{2m}$$ can be viewed purely as permutations of the ($$m$$ many) vertices. In particular, for $$m \leq n$$, $$D_{2m}$$ can be viewed as a symmetry of $$n$$ points: Just pick $$n - m$$ points to be fixed, and then let $$D_{2m}$$ act on the remaining $$m$$ points by identifying them with the vertices of an $$m$$-gon. Of course, this is exactly your idea!

This shows that $$D_{2m}$$ is actually a subgroup of $$S_n$$ in multiple ways. After all, we have a choice of which $$n-m$$ points to be fixed, and we also have a choice of "initial configuration" of our $$m$$ points on the polygon (as a cute exercise, can you figure out exactly how many copies of $$D_{2m}$$ this gives us inside $$S_n$$ for $$n \geq m$$?).

Ok, now for some stuff not directly related to your question:

Intuitively, these copies of $$D_{2m}$$ are kind of the same, right? Like, they're all the same action of $$D_{2m}$$ on $$n$$ many points. The only distinction is in how we choose the "initial configuration". When $$D_8 \leq S_5$$, do we want to leave $$1$$ fixed? Or $$2$$? It doesn't really matter, right?

The way we make this precise is with the notion of conjugate subgroups. We say two subgroups $$H, H' \leq G$$ are conjugate if there is some $$g \in G$$ with $$g^{-1}Hg = H'$$. Obviously if $$H$$ and $$H'$$ are conjugate, then we must have $$H \cong H'$$ (do you see why?). Importantly, though, the converse is false! It's possible to have $$H \cong H'$$ both subgroups of $$G$$, but with $$H$$ not conjugate to $$H'$$! See here, for instance.

You should think of isomorphic, nonconjugate subgroups as being "accidentally" the same kind of symmetry, whereas conjugate subgroups "have a good reason" to be isomorphic -- because they're really representing the same group action, just in different ways! We saw this above with the copies of $$D_{2m}$$ in $$S_n$$. They're really the same action.

For a simple example, consider the following $$3$$ subgroups inside $$D_8$$:

The two copies of $$\mathbb{Z}/2$$ given by the flips in the top row are conjugate. They're really "the same flip", up to a rotation of the square. Contrast this with the copy of $$\mathbb{Z}/2$$ given by the diagonal flip underneath. Even though this is the same subgroup up to isomorphism, it arises in an entirely different way from those in the top row.

As an interesting pathology, you can actually have $$H \cong H' \leq G$$, with $$G / H \not \cong G / H'$$! See here for an example. Of course, if $$H$$ and $$H'$$ are conjugate, then this doesn't happen.

The difference between "isomorphic" and "conjugate" is a subtle but important one, and quite a bit of more advanced group theory rests on this concept. I know this is a bit far removed from your original question, but it felt more interesting to chat about this then to just leave a comment saying "yes, you're right" :P

I hope this helps ^_^

Yes. A geometry-free argument goes as follows. Take any subgroup $$H$$ of order $$2$$ of $$D_{2m}$$; now, $$D_{2m}$$ acts by left multiplication on the left quotient $$D_{2m}/H$$ (abuse of notation), which has cardinality $$m$$; the kernel of this action is the normal core of $$H$$ in $$D_{2m}$$, which is trivial (think of, e.g, the intersection between $$H=\{1,s\}$$ and $$rHr^{-1}=\{1,r^{2}s\}$$, for $$m>2$$). Therefore, $$D_{2m}\hookrightarrow S_m$$. In turn, for $$n>m$$, $$S_m\hookrightarrow S_n$$ in the obvious way (extension by the identity map of the set $$\{m+1,\dots,n\}$$). Finally, consider the composition of the two embeddings, and you end up with an embedding of $$D_{2m}$$ into $$S_n$$.