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?


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 $H=K$.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.