How many normal subgroups is in a free group of rank > 1

Question

How many normal subgroups is in a free group of rank $k>1$, if the quotient group (of the normal subgroups) isomorphic $S_3$?

Answer 1

Following Jim Belk's helpful comment, I make it $(6^k-3^k-3(2^k-1))/6$.

Summary of proof. Each map from the set of free generators of $F_k$ to $S_3$ defines a homomorphism, and so there exactly $6^k$ such homomrophisms. We want to exclude those whose image is a proper subgroup of $S_3$. Now $3^k$ of them map onto the subgroup of order $3$ and $2^k$ map onto each of the three subgroups of order $2$, so the total number of surjective homomorphisms $F_k \to S_3$ is $6^k-3^k-3(2^k-1)$. (It is $2^k-1$ because we have already discounted the trivial map.)

But if you compose a homomorphism with any of the $6$ automorphisms of $S_3$, then you get a homomorphism with the same kernel, and conversely, two epimorphisms have the same kernel if and only one is the other composed with an automorphism of $S_3$. So we divide by $6$ to get the number of normal subgroups of $F_k$ with quotient $S_3$.