# Showing that $H\leq S_n$ containing rotations must be isomorphic to $D_n$

Let $$H$$ be a subgroup of $$S_n$$ such that $$H$$ is isomorphic to the dihedral group $$D_n$$. Let also $$K$$ be a subgroup of $$S_n$$ such that $$K$$ is isomorphic of $$D_n$$.

I would like to show that if $$K$$ contains all the rotations from $$H$$, then $$K= H$$.

I have verified this for $$n=5$$. For example, if $$H = \{(1), r, r^2, r^3, r^4, a,b,c,d,e\}$$, where $$r$$ is the specific rotation $$r= (1 2 3 4 5)$$ and $$a,b,c,d,r$$ are the rotations (elements of order $$2$$). Then I have verified by brute force, that should another subgroup $$K$$ contain all the rotations $$(1),r,r^2,r^3,r^4$$ (and $$K\simeq D_5$$), then $$H = K$$ (so they are actually equal).

My question is: How can I find a general argument for this?

## 2 Answers

It is not true that if K contains all the rotations from H, then K=H.

Take $$n=12$$ and consider the cyclic group $$\langle \alpha \rangle$$, where $$\alpha=(1234)(567)$$. Set $$\beta_1=(24)(56)$$. Then $$H:=\langle \alpha,\beta_1 \rangle\cong D_{12}$$, since $$\alpha$$ has order $$12$$, $$\beta_1$$ has order $$2$$ and $$\beta_1\alpha=\alpha^{11}\beta_1$$.

But for $$\beta_2=(24)(56)(89)$$ we also have that $$\alpha$$ has order $$12$$, $$\beta_2$$ has order $$2$$ and $$\beta_2\alpha=\alpha^{11}\beta_2$$, hence $$K:=\langle \alpha,\beta_2 \rangle\cong D_{12}$$, but evidently $$H\ne K$$.

On the other hand, if the cyclic $$n$$-group inside $$H$$ is generated by an $$n$$-cycle, the result is true, by the argument given by @Andrea Mori.

In that case let $$\alpha=(1\ 2 \dots n)$$ be the generating $$n$$-cycle. There are $$n$$ involutions $$\beta_1,\dots \beta_n$$ in $$S_n$$ that satisfy $$\beta_k\alpha=\alpha^{n-1}\beta_k$$. In fact, set $$\beta_k(j)=n+k+1-j,$$ then it is straightforward to show that $$\beta_k\circ \beta_k=Id$$ and that $$\beta_k\alpha=\alpha^{n-1}\beta_k$$. Any of them can be chosen to generate $$D_n$$, and then the others are in the same group.

On the other hand, assume that $$K\subset S_n$$ shares the same rotation group as $$H$$ and $$K\cong D_n$$.

Take any involution $$\beta$$ of $$K$$ and set $$k=\beta(1)$$. We will show that $$\beta=\beta_k$$. Clearly $$\beta(1)=k=\beta_k(1)$$.

Now one uses that $$\alpha \beta \alpha=\beta$$ in order to show inductively that $$\beta(j)=n+k+1-j=\beta_k(j)$$ for all $$j=n,n-1,\dots,2$$ (computing all numbers modulo $$n$$).

We begin with $$\beta(n)=\alpha(\beta(\alpha(n)))=\alpha(\beta(1))=\alpha(k)=k+1=\beta_k(n).$$

Then we compute $$\beta(n-1)=\alpha(\beta(\alpha(n-1)))=\alpha(\beta(n))=\alpha(k+1)=k+2=\beta_k(n-1).$$ In general, if we have already $$\beta(j)=n+k+1-j$$, then we compute $$\beta(j-1)=\alpha(\beta(\alpha(j-1)))=\alpha(\beta(j))=\alpha(n+k+1-j)=n+k+1-(j-1)=\beta_k(j-1),$$ which shows that $$\beta=\beta_k$$, hence $$H=K$$.

• The example for $D_6$ was incorrect, since in that case $H=K$. – san Oct 31 '18 at 1:56
• San, thanks for your answer. Would you be able to add more details to Andrea's argument? – John Doe Oct 31 '18 at 16:27

Up to relabelling the vertices of the polygon you may assume that the cyclic group of rotations in $$K=H$$ is generated by $$(1\ 2\ \cdots n)$$.

So you can now reason geometrically: if both $$K$$ and $$H$$ are dyhedral groups, i.e. groups of isometries of the regular $$n$$-gon, their involutions must shuffle the same pairs of vertices. Thus $$H=K$$ because for both of them the vertices have the same labelling.

In other words: the only way to embed $$D_n$$ into $$\cal S_n$$ is via some labelling of the vertices of the $$n$$-gon (up to an obvious equivalence) and this completely characterizes the image of the embedding.

• Thanks for the answer. I am still not sure about the " must shuffle the same pairs of vertices. " part. – John Doe Oct 22 '18 at 14:32