# Parity of order of intersection of cyclic and noncyclic subgroups

The dicyclic group with parameter $n$, denoted by ${Dic} \left(n \right)$, is defined as \begin{equation} {Dic} \left(n \right) \mathrel{\mathop:}= \langle a,b \mid a^{2n} = e, b^2 = a^n, bab^{-1} = a^{-1} \rangle \end{equation}

I want to prove the following results:

Let $a$ be the generator of Dic(n) with $\vert \langle a \rangle \vert = 2n$. Let $H$ be a subgroup of ${Dic} \left(n \right)$. Then either

1.) $H \subseteq \langle a \rangle$ or

2.) For some $k \mid n$, $\vert H \cap \langle a \rangle \vert = 2k$ and $\vert H \vert = 4k$.

Proof:

Note that $\langle a \rangle \trianglelefteq {Dic} \left(n \right)$. Hence $H \cdot \langle a \rangle \leq {Dic} \left(n \right)$ with $\vert H \cdot \langle a \rangle \vert \bigm| 4n$. Thus

$$\vert H \cdot \langle a \rangle \vert = \frac{\vert H \vert \cdot \vert \langle a \rangle \vert}{\vert H \cap \langle a \rangle \vert} = \frac{\vert H \vert \cdot 2n}{\vert H \cap \langle a \rangle \vert}$$ Then $\frac{\vert H \vert}{\vert H \cap \langle a \rangle \vert} \bigm| 2$ and so either $\frac{\vert H \vert}{\vert H \cap \langle a \rangle \vert} = 1$ or $\frac{\vert H \vert}{\vert H \cap \langle a \rangle \vert} =2$.

The first case shows that $H \subseteq \langle a \rangle$. For the second case, we have $\vert H \vert = 2 \vert H \cap \langle a \rangle \vert$. I want to show that $\vert H \cap \langle a \rangle \vert = 2k$ for some $k$, i.e, the order is always even. Is it possible to show this? I've tried solving $H \cap \langle a \rangle$ of some dicyclic groups and indeed it is even. Also, when the intersection is odd, $H$ is actually not a subgroup. Is there a way to prove that $\vert H \cap \langle a \rangle \vert$ is always even?

I claim that $a^n\in H\cap\langle a\rangle$. Take $x\in H\setminus H\cap\langle a\rangle$, which is of the form $a^j b$ for some $0\le j\le 2n-1$. Then $$(a^jb)^2 =a^ja^{-j}b^2 = a^n$$ so $a^n\in H\cap\langle a\rangle$
If $a^k\in H\cap\langle a\rangle$ and $0\le k < n$ then $a^{n+k}\in H\cap\langle a\rangle$, so we can correspond $a^k$ to $a^{n+k}$ and vice versa. Hence $|H\cap\langle a\rangle|$ is even.