Solution: Recall that \begin{align*} & i^2 = j^2 = k^2 = -1,\\ & ij=k,jk=i,~\text{and}~ki=j,\\ & i^4 = j^4 = k^4 = 1. \end{align*} Thus, we can write $Q_8$ as \[ \langle i,j \rangle = \langle j,k \rangle = \langle k,i \rangle. \]

We can now solve the problem by listing all the cyclic subgroups, namely \begin{align*} & \left\langle i \right\rangle = \langle i,-1,1 \rangle ,~~ \langle j \rangle = \langle j,-1,1 \rangle ,~~ \langle k \rangle = \langle k,-1,1 \rangle \\ & \langle -1 \rangle = \{-1,1\}~~\text{and}~~\langle 1 \rangle = \{1\} \end{align*} Since, the quaternion group has order $8$ and hence using the Lagrange's Theorem, the possible order of subgroups are $1,2,4$ and $8$. Since the group is not cyclic, we will not have any element of order $8$. Therefore we have \begin{align*} & 1 ~~\text{element of of order}~~ 1, \\ & 1 ~~\text{element of of order}~~ 2, \text{ and } \\ & 6 ~~\text{element of of order}~~ 4. \end{align*}

The following theorem determines the number of cyclic subgroup of order $k$ of a finite group

Let $G$ be a finite group. Then the number of cyclic subgroups of order $k$ is $G$ is \begin{equation*} \dfrac{\text{number of elements of order $k$ in $G$}}{\phi(k)}, \end{equation*} where $\phi(k)$ is the Euler's phi function which is the number of positive integers less than $k$ that are relatively-prime to $k$.

Now we will use the above theorem to find number of cyclic subgroups of order $1,2$ and $4$.

1. Number of cyclic subgroups of order $1:$ \begin{align*} & = \dfrac{\text{number of elements of order $1$ in $G$}}{\phi(1)} = \frac{1}{1} = 1. \end{align*}
2. Number of cyclic subgroups of order $2:$ \begin{align*} & = \dfrac{\text{number of elements of order $2$ in $G$}}{\phi(2)} = \frac{1}{1} = 1. \end{align*}
3. Number of cyclic subgroups of order $4:$ \begin{align*} & = \dfrac{\text{number of elements of order $4$ in $G$}}{\phi(4)} = \frac{6}{2} = 3. \end{align*}

Thus, the total number of cyclic subgroups of $Q_8$ is $1 + 1 + 3 = 5$.