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

Question

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?

Answer 1

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 $H=K$.

Answer 2

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.