Suppose $D_n$ is the dihedral group of order $2n$ ($D_n = \{1, a, a^2, \ldots, a^{n-1},b,ab,\ldots, a^{n-1}b\}$ with $a^n = 1, b^2 = 1$ and $ba = a^{-1}b$).

I have shown that if $H \leq \{1, a,\ldots, a^{n-1}\}$, then $xH = Hx$ for all $x \in D_n$.

The follow-up question asks to prove that if $K$ is a non-abelian subgroup of $D_n$, and $H = K \cap \{1, a,\ldots, a^{n-1}\}$, then the index of $H$ in $K$ is 2.

This question comes in the section where cosets are introduced, so I feel that some 'basic' reasoning should be possible. However, I can not figure it out. I tried using the previous part, but could not proceed. I know that since $K$ is non-abelian and $H$ is, the index is at least 2. I tried proving that it can not be larger than 2, but got stuck.

In another approach, I got that $K$ contains some $a^kb$ and tried proving that $H$ and $a^kb$ are the only cosets in $K$, but got stuck again...

Any hints would be appreciated...

  • $\begingroup$ When you write $H\subset \{1,a,\ldots,a^{n-1}\}$, I think you mean $H$ is a subgroup, rather than a subset. It's preferable to use the symbol $\leq$. $\endgroup$ – verret Jul 25 at 1:33
  • $\begingroup$ @verret you're right, I have edited it. $\endgroup$ – Student Jul 25 at 5:36

Note: I have treated two extreme cases separately (even though there is no need of that) just for the sake of clarity.

If $a \in K$, then $H=\langle a \rangle \subset K$. In which case $[K:H]=\frac{|K|}{|H|}=\frac{|K|}{n}$. But $|K| \leq 2n$, so index is either $1$ or $2$. But $H \neq K$ (since $K$ is non-abelian and $H$ is abelian), so $[K:H]=2$.

If $K \cap \langle a \rangle =\{1\}$. Then $|K|=2$ (only reflections and no rotation). In which case the index is $2$.

The last scenario is when $1<i<n-1$ is the smallest exponent such that $a^i \in K$. In which case $H=\langle a^i\rangle$. Since $K$ is non-abelian, some reflection is in $K$. In which case $K=\langle a^i, a^jb\rangle$. Consequently $K=H \cup (a^jb)H$, hence the index is $2$.


The statement of your question is a consequence of the fact that the subgroups of a dihedral group are either cyclic or dihedral themselves (see the proof of this here). So the group $K$, being non abelian, is necessarily dihedral. It is not hard to see that $H$ is the maximal cyclic subgroup of his big brother $K$, and the latter being dihedral makes $H$ have index $2$.


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