Number of normal subgroups from $F_{2}$ which factor groups are isomorphic to $D_{n}$

Question

What is the number of normal subgroups of the free group $F_{2}$ whose factor groups are isomorphic to the dihedral group $D_{n}$?

Answer 1

The number is actually 3 (provided $n>2$): let $a,b$ be free generators of $F=F_2$ then we can distinguish what happens to them when mapped by a surjective morphism $F_2\twoheadrightarrow D_n$:

1. $a$ is mapped to a rotation, $b$ to a reflection (as mentioned in the other answer, you have several choices for such morphisms, but the kernel is the same for all these choices): the kernel is the normal subgroup generated by $\{a^n,b^2,(ab)^2\}$

2. $b$ is mapped to a rotation, $a$ to a reflection (again there are several morphisms of this kind but only one normal subgroup): the kernel is the normal subgroup generated by $\{a^2,b^n,(ab)^2\}$

3. $a$ and $b$ are both mapped to (distinct) reflections (the order of the product of these has to be $n$ --- again there are several morphisms of this kind, but only one normal subgroup): the kernel is the normal subgroup generated by $\{a^2,b^2, (ab)^n\}$.

It is not hard to see that these three are indeed pairwise distinct (in particular, the first two of them are distinct, so the number is not $2$).

Answer 2

In other words: How many epimorphisms $F_2\to D_n=\langle\,r,s\mid\,r^n=s^2=srsr=1\,\rangle$ are there?

If $a,b$ are the images of the generators of $F_2$ under such a homomoprhism, then clearly $D_n=\langle a,b\rangle$ and our normal subgroup, i.e., the kernel, is uniquely determined by the choice of generators $a$ and $b$ of $D_n$.

To avoid that $\langle a,b\rangle$ consists only of rotations, at least one of $a,b$ must be a reflection (conjugate of $s$). Then the other of the two or their product $ab$ must be a rotation $r^k$. To generate the whole group, we need that $k$ is coprime to $n$. So we can count these cases:

• $a$ is one of $n$ reflections; $b$ is one of $\phi(n)$ rotations
• $b$ is one of $n$ reflections; $a$ is one of $\phi(n)$ rotations
• $a$ is one of $n$ reflections; $b$ is $a$ times one of $\phi(n)$ rotations

In total we find $3n\phi(n)$ pairs $(a,b)$. By the above remarks, $$3n\phi(n)$$ is also the answer to the problem.