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I'll use a concrete definition of a dihedral group $D_{2n}$ which emphasizes its group structure:

$D_{2n}$ consists of distinct elements $r_0,...,r_{n-1},s_0,...,s_{n-1}$ so that for any $i \in \mathbb{Z}$ we have $r_{i + n} = r_i, s_{i + n} s_i$ and for any $i,j \in \mathbb{Z}$ we have $r_ir_j = r_{i + j}, r_is_j = s_{i + j}, s_ir_j = s_{i - j}, s_is_j = r_{i - j}$.

It's easy to see that all $r_i's$ commute in $D_{2n}$, hence the only elements for which it is possible not to commute are $s_i$ and $s_j$ or $r_i$ and $s_j$.

$s_is_j = s_js_i$ whenever $i - j \equiv j - i \bmod n$ and $r_is_j = s_jr_i$ whenever $i + j \equiv i - j \bmod n$, hence $D_{2n}$ is noncommutative if and only if either

  • there is $0 \leq j < n$ so that $n\nmid 2j$, or

  • there are $0 \leq i,j < n$ so that $n\nmid 2(i - j)$.

Using this result, it's not hard to establish that $D_2$ and $D_4$ are abelian.

I wonder how we can use this to prove that $D_{2n}$ is not abelian for $n \geq 3$. I'm not interested in a possible geometric proof.

Jxt921
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2 Answers2

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I bet you can prove that $$r_1s_1\neq s_1r_1$$ It follows easily from your definition.

Matt Samuel
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By definition there is a rotation $r$ in $D_n$, $n\ge 3$ and a reflection $s$ such that $$ srs^{-1}=r^{-1}\neq r. $$ So $D_n$ is non-abelian. We can also give a second proof by determining the center of $D_n$. A group is abelian iff it equals its center. However, we easily see that $Z(D_n)$ is much smaller than that:

The center of the dihedral group

Dietrich Burde
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