The dihedral group consists of rotations and symmetries. But the symmetry group is a group only if n is even, thus the group of rotations is a normal subgroup of the dihedral group.

So how to describe this normal subgroup(rotations) and are there any other normal subgroups of the dihedral group?

  • $\begingroup$ Can you explain what you mean by "the symmetry group is a group only if n is even"? $\endgroup$ Dec 6, 2013 at 19:41
  • $\begingroup$ By "the symmetry group", do you mean "the symmetric group", i.e. the group of permutations of $n$ things? Usual language is that every object defines a "symmetry group". The _$n$th symmetric group_ $S_n$ is then the symmetry group of a set of size $n$. The _$n$th dihedral group_ $D_n$ is the symmetry group of a regular $n$-gon. $\endgroup$ Dec 6, 2013 at 19:47

1 Answer 1


I will denote by $C_n < D_n$ the subgroup of rotations. Note a few things: $C_n$ has index $2$ in $D_n$, hence is normal, as you point out. Moreover, $D_n$ is a semidirect product $S_2 \ltimes C_n$, where $S_2$ acts on $C_n$ by $x \mapsto x^{-1}$. Here $S_2$ is the group of order $2$. Note also that $C_n$ is cyclic: it is generated by a rotation by $2\pi/n$.

Let us first classify which subgroups $H$ of $C_n$ are normal in $D_n$. Let $y\in D_n$. Then either $y \in C_n$ or $y = xs$ for $x\in C_n$ and $s=s^{-1}$ is some particular choice of reflection. For $h\in H$, we see that if $y\in C_n$, then $yhy^{-1} = h$; if $y = xs$ for $x\in C_n$, then $yhy^{-1} = xshsx^{-1} = xh^{-1}x = h^{-1}$. In either case, $yhy^{-1} \in H$. Thus every subgroup of $C_n$ is normal in $D_n$. The subgroups of $C_n$ are classified by numbers $d$ dividing $n$.

Second, is it possible for a normal subgroup of $D_n$ to fail to be contained in $C_n$? Yes, of course: $D_n$ is normal in itself. Suppose that $H < D_n$ is normal and $s\in H$ for some reflection $s$. There are two cases, depending on the parity of $n$:

If $n$ is odd, then all reflections are in the same conjugacy class. The generator of $C_n$ ("rotation by one click") is a product of two "adjacent" reflections. Thus if $n$ is odd, then $H > C_n$, and since $C_n$ has index $2$, $H = D_n$.

If $n$ is even, then there are two conjugacy classes of reflections: those that go through vertices of the regular $n$-gon, and those that go through edges. These are equivalent under an outer automorphism of $D_n$. Suppose $H < D_n$ is normal and $s \in H$ is a reflection. Let $c$ denote the generator of $C_n$ ("rotation by one click"); then $csc^{-1}\in H$, and $csc^{-1}s = c^2$. Thus $H$ contains the subgroup of $C_n$ generated by $c^2$. If $H \ni c$, then $H = D_n$; otherwise, $H$ is generated by $s$ and $c^2$. It is therefore a copy of $D_{n/2}$ living in $D_n$, and hence has index $2$; it is therefore normal.

In summary, the normal subgroups of $D_n$ are:

  • $D_n$ itself.
  • Any subgroup of $C_n$. Theses are in bijection with the positive integers $d$ dividing $n$.
  • If $n$ is even, then there are two more normal subgroups of $D_n$, each isomorphic to $D_{n/2}$. Both contain the subgroup $C_{n/2} < C_n$. The remaining $n/2$ elements are one or the other of the two conjugacy classes of reflections in $D_n$.

Not the answer you're looking for? Browse other questions tagged .