Dihedral Group is Non-Abelian

Theorem

Let $n \in \N$ be a natural number such that $n > 2$.

Let $D_n$ denote the dihedral group of order $2 n$.

Then $D_n$ is not abelian.

Proof

From Group of Order less than 6 is Abelian we have that $D_1$ and $D_2$ are abelian, which is why the condition on $n$.

From Group Presentation of Dihedral Group we have:

$\beta \alpha = \alpha^{n - 1} \beta$

for some $\alpha, \beta \in D_n$ such that $\alpha \ne \beta$.

We also have:

$\alpha^n = e$

But if $D_n$ were abelian, that would mean:

$\beta \alpha = \beta \alpha^{n - 1}$

and so by the Cancellation Laws:

$\alpha = \alpha^{n - 1}$

and so:

$\alpha^2 = e$

But by definition of the dihedral group, this is not the case.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 29 \gamma$