Quaternion Group is Hamiltonian

Theorem
The quaternion group $Q$ is Hamiltonian.

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

By definition $Q$ is Hamiltonian :
 * $Q$ 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:


 * $Q$
 * $\gen {\mathbf i}$, $\gen {\mathbf j}$, $\gen {\mathbf k}$
 * $\gen {-\mathbf 1}$
 * $\set {\mathbf 1}$

From Trivial Subgroup and Group Itself are Normal:
 * $Q$ and $\set {\mathbf 1}$ are normal subgroups of $Q$.

$\gen {-\mathbf 1}$ 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 $\gen x$ generated by $x$.

The order of element $x$ is $4$:
 * $x^4 = \paren {x^2}^2 = \paren {-\mathbf 1}^2 = 1$

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.

The 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.