# Let K be a non-abelian subgroup of dihedral group and $H = K \cap \{1, a, a^2, \ldots, a^{n-1}\}$ then the index of H in K is 2.

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...

• 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 $\leq$. – verret Jul 25 at 1:33
• @verret you're right, I have edited it. – Student Jul 25 at 5:36

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 $$1 is the smallest exponent such that $$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$$.