Quaternion Group not Dihedral Group

Theorem
Let $Q$ be the quaternion group.

Then $Q$ is not isomorphic to the dihedral group $D_4$.

Proof
From Group Presentation of Dihedral Group, $D_4$ is generated by two elements of orders $4$ and $2$ respectively.

Let these generators be $\alpha$ and $\beta$ where:
 * $\alpha^4 = e$
 * $\beta^2 = e$

Hence $\alpha^2$ and $\beta$ are different elements of $D_4$ which both have order $2$.

But the quaternion group:
 * $Q = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$

has only one element of order $2$, which is $-\mathbf 1$.

The rest have order one or four.

Thus $Q$ and $D_4$ cannot be isomorphic because there are not the same number of elements of order $2$ in both groups.