Quaternion Group is Hamiltonian

Theorem
The quaternion group $Q$ is Hamiltonian.

Proof
For clarity the Cayley table of $Q$ is presented below:


 * $\begin{array}{r|rrrrrrrr}

& \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\ \hline \mathbf 1 & \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\ \mathbf i & \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf 1 &  \mathbf k & -\mathbf j & -\mathbf k &  \mathbf j \\ -\mathbf 1 & -\mathbf 1 & -\mathbf i & \mathbf 1 &  \mathbf i & -\mathbf j & -\mathbf k &  \mathbf j &  \mathbf k \\ -\mathbf i & -\mathbf i & \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf k &  \mathbf j &  \mathbf k & -\mathbf j \\ \mathbf j & \mathbf j & -\mathbf k & -\mathbf j &  \mathbf k & -\mathbf 1 &  \mathbf i &  \mathbf 1 & -\mathbf i \\ \mathbf k & \mathbf k &  \mathbf j & -\mathbf k & -\mathbf j & -\mathbf i & -\mathbf 1 &  \mathbf i &  \mathbf 1 \\ -\mathbf j & -\mathbf j & \mathbf k &  \mathbf j & -\mathbf k &  \mathbf 1 & -\mathbf i & -\mathbf 1 &  \mathbf i \\ -\mathbf k & -\mathbf k & -\mathbf j & \mathbf k &  \mathbf j &  \mathbf i &  \mathbf 1 & -\mathbf i & -\mathbf 1 \end{array}$

By definition $Q$ is Hamiltonian iff $Q$ is non-abelian and every subgroup of $Q$ is normal.

$Q$ is non-abelian as demonstrated by the counter-example:


 * $\mathbf{ij} \ne \mathbf{ji}$

Using Lagrange's theorem and inspection it can be established that the subgroups of $Q$ are:


 * $Q$
 * $\left\langle {\mathbf i} \right\rangle$, $\left\langle {\mathbf j} \right\rangle$, $\left\langle {\mathbf k} \right\rangle$
 * $\left\langle {-\mathbf 1} \right\rangle$
 * $\{\mathbf 1\}$

From Trivial Subgroup and Group Itself are Normal, $Q$ and $\{1\}$ are normal subgroups of $Q$.

$\left\langle {-\mathbf 1} \right\rangle$ is the center of $Q$ and therefore normal.

Let $x$ denote one of $\mathbf i$, $\mathbf j$ or $\mathbf k$.

Let $X$ denote cyclic subgroup $\left\langle {x} \right\rangle$ generated by $x$.

Order of element $x$ is $4$:
 * $x^4 = \left(x^2\right)^2 = \left(-\mathbf 1\right)^2 = 1$

which can be seen by exploiting properties of the above table.

This means that order of $X$ equals $4$.

Every coset space of $X$ in $Q$ must consist of:
 * $X$ itself by Coset by Identity,
 * its complement $Q \setminus X$ by Cosets are Equivalent and Coset Spaces form Partition.

Coset space of $X$ in $Q$ has two elements, so $X$ have an index of $2$.

From Subgroup of Index 2 is Normal it follows that all subgroups of $Q$ are normal.

Hence the result.