# Quaternion Group not Dihedral Group

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## 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.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \lambda$