Prove that if $k\mid n$ that there exist subgroup $H\le D_n$ such that $H \cong D_k$.

Question

Prove that if $k$ divides $n$ that there exists a subgroup $H \le D_n$ such that $H \cong D_k$. Provide an example in which $H$ is not a normal subgroup of $D_n$.

This is the problem that I am looking at. Here is what I have so far.

Part a)

Let $D_n$ denote the dihedral group of order $2n$, which consists of the symmetries of a regular $n$-gon. Let $k$ divide $n$, i.e., $n = km$ for some integer $m$. Then, we can consider the subgroup $H$ of $D_n$ consisting of the symmetries of a regular $k$-gon inscribed in the regular $n$-gon. Specifically, $H$ consists of rotations of the regular $n$-gon by multiples of $\frac{2\pi}{k}$ radians, and reflections across lines passing through vertices of the regular $k$-gon.

It is clear that $H$ has order $2k$ ($k$ rotations and $k$ reflections), which divides the order of $D_n$, which is $2n$. To show that H is isomorphic to $D_k$, we need to find a bijection (i.e., one-to-one and onto function) between the elements of $H$ and $D_k$ that preserves the group operation.

One such bijection is to send a rotation of $\frac{2\pi}{k}$ radians to the corresponding rotation of $\frac{2\pi}{k}$ radians in $D_k$, and to send a reflection across a line passing through a vertex of the regular $k$-gon to the corresponding reflection in $D_k$. This bijection preserves the group operation because rotations and reflections in $D_k$ satisfy the same relations as the corresponding elements in $H$.

Part b)

Now an example where $H$ is not a normal subgroup of $D_n$. Let $n = 6$ and $k = 2$, so that $H$ is the subgroup of $D_6$ consisting of the identity, a rotation of $\pi$ radians, and three reflections across lines passing through vertices of an inscribed regular triangle. Note that $H$ is isomorphic to $D_2$, which is the cyclic group of order $2$.

Consider the element $\sigma$, a reflection across a line passing through two non-adjacent vertices of the regular hexagon, which is not in $H$. Then, the conjugate of $\sigma$ by a rotation of $\frac{2\pi}{6}$ radians is another reflection that is not in $H$. Therefore, $H$ is not a normal subgroup of $D_6$.

Is this sufficient?

Answer 1

We know that $D_{n}\cong \mathbb{Z}_{n} \rtimes \mathbb{Z_{2}}$, where $\mathbb{Z_{2}}=\langle x\rangle$ and acts by inverse on $\mathbb{Z_{n}}$ (Hence acts by inverse on any subgroup of $\mathbb{Z_{n}}$.), meaning that for every element $g\in Z_n$, $g^x=g^{-1}$.
Here we're looking for $D_{k}\cong \mathbb{Z}_k \rtimes \mathbb{Z_{2}}$. As we know, $\mathbb{Z_{n}}$ has a unique subgroup isomorphic to $\mathbb{Z}_{k}$, say $K$. Now by the discussion we had, we get that $D_{k}\cong K\rtimes \mathbb{Z_{2}}\le D_{n}$.

Answer 2

You can use $D_8=\Bbb Z_4\rtimes \Bbb Z_2$ for the second part.

There's $D_2$'s, generated by reflections, that are not normal. Because $\rho f=f\rho^{-1}\implies \rho f\rho^{-1}=f\rho^{-2}.$

Of course if the $\Bbb Z_2$ on the right side was normal we would just have a direct product. Of course you can also do $D_6,$ by the same reasoning. Or any larger $D_n.$

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Actually, there are going to be $n/k$ subgroups of type $D_k$.

So say $n=12.$ There's three subgroups of type $D_4.$ Thus by Sylow they are not normal.

This will always happen $k$ is the highest power of $2$ dividing $n$ (and $k\lt n$).