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A geometric proof that D2mSn is possible for m>n

16 Aug 2021

To nobody’s surprise, I was on MSE tonight, and saw a simple question about group theory. The original question doesn’t matter as much as a question it made me wonder to myself:

Is there an m>n so that the dihedral group D2m is a subgroup of the symmetric group Sn?

Intuitively it feels like the answer should be “yes”, but I wasn’t able to come up with a proof myself. Thankfully it didn’t take much googling to find an excellent example due to pjs36. I’ll show it for completeness, but you can find the original post here.

The idea is to embed D12 in S5 by working with subpolygons instead of vertices. This is analogous to showing the symmetry group of a cube is S4 by looking at the diagonals inside the cube, rather than looking at the vertices/edges/faces individually.

Since a picture is worth a thousand words, I’ll steal pjs36’s picture:

If you know where these 5 subpolygons get sent, you actually know where the whole hexagon gets sent! This witnesses D12S5 in a starkly beautiful way.

It got me wondering, though. Can we run a similar argument to get D2mSn for other choices of m? Since you’ve read the title of this blog post, you know the answer is “yes”1. But at this point I should issue a quick clarification: This (very clever!) idea is not actually my own – I found it in yet another mse post. I was already planning on writing up a blog post about this problem, but when I found the solution I knew I had to talk about it. Now, let’s talk through how we might have solved this problem ourselves:

Obviously if p is prime, then D2p first shows up as a subgroup of Sp in the natural way (by permuting the vertices). We can’t do any better than this since for p prime, pn! for any n<p.

A moment’s thought (or more likely, quite a few moments’ thoughts) shows that actually D2pkSn for n<pk. The idea here is that D2pk has a pk cycle. The order of an element σSn is the lcm of the cycle lengths in σ, so even though pk might divide n!, there’s no way to get an lcm of things <pk to equal pk.

Pjs36’s solution feels like it should generalize, and looking more carefully at the pictures above, we’re considering the 2-gons and 3-gons living inside of our 6-gon… This idea links up well with our counterexamples from earlier, since the subpolygons of an m-gon are exactly the -gons for m and prime powers are special in the divisibility order.

We want to make n (the number of generators of our symmetric group) as small as possible, so we should make the subpolygons we look at as big as possible (so that there aren’t many). Since we know from our earlier investigation that pk-gons are the obstruction to “shrinking” n, whatever construction we do should give us pk objects to permute when we look at a pk-gon.

Eventually, this might lead us to consider the pk many mpk-gons living inside our m-gon. In the pk-gon case, this means we’re considering the pk many 1-gons (that is, just the vertices), which is exactly what we expect. In the 6-gon case, this mean we consider the 62-gons and the 63-gons, but this is exactly pjs36’s original example! In the case of a 23325-gon, this means we’re looking at the 45-gons (there’s 8 of them), the 40-gons (there’s 9 of them), and the 72-gons (there’s 5 of them). Notice here how we’re able to keep the number of objects small (there’s only 8+9+5=22 things we’re permuting) by keeping our subpolygons big.

In general, let’s write m=Pi where each Pi=piki is a (maximal) prime power. We can find Pi many mPi-gons living inside our m-gon, and every symmetry of our m-gon permutes these subpolygons amongst themselves. That is, we get a permutation in SPi for each i.

Next, we can glue these together into a map

D2mSPi.

which I claim is actually injective.

To see why, say that some gD2m is in the kernel of the above map. Then it’s in the kernel of each D2mSPi, so g fixes each of our subpolygons. But this can only happen if g fixes the entire m-gon, so g=1 and our map is injective.

Now we see the light at the end of the tunnel! We have an embedding D2mSPi, and we want an embedding D2mSn for some n. But there’s an “obvious” way to do this! If you have a permutation of k objects and a permutation of objects, we can just put them next to each other and call it a permutation of k+ objects.

Using this “put them next to each other” embedding, we see that

D2mSPiSPi.

So D12 embeds in S2+3=S5, as we’ve seen. Likewise, each D2pk embeds in Spk, which agrees with our earlier experiments. Finishing our concrete example from earlier shows D2(23325) embeds in S23+32+5=S22, which is much smaller than the obvious S23325=S360!

As one last question: how much smaller is it? If m=piki, then let’s call each piki a Principal Divisor of m. Moreover, let’s write

s(m)=piki.

According to Upper Bounds on the Sum of Principal Divisors of an Integer by Eggleton and Galvin (up to change of variable name, to be consistent with the other variables in this post):

If m is any positive integer with 2 principal divisors, and each greater than /2, then

s(m)m2.

Moreover, this holds with equality when m=30.

This tells us that we can embed D2m in a symmetric group with generators that shrink rapidly as the number of principal divisors of m increases (provided each is not individually too small).

We’ve shown that D2mSs(m), but maybe we can do even better!

As a (fun?) exercise, can you show that that isn’t the case? That is, if D2m embeds in Sn, show that ns(m), so our construction here was best possible.

The proof is similar to how we showed D2pk can’t embed in Sn for n<pk.


  1. This is actually a special case of a very hard problem. In general for a group G we have very little idea what the minimal n with GSn is. It’s super cool that we can solve this explicitly for dihedral groups!