# Nonabelian dihedral groups and a question in number theory [duplicate]

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.

## marked as duplicate by Dietrich Burde group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 13 at 19:46

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