What is the number of normal subgroups of the free group of rank 2 $F_{2}$ whose factor groups are isomorphic to the $Q_{8}$?
I have a plan solutions. But I can't get a numerical answer.
- Minimum generating $Q_{8}$ system consists of $2$ elements, because it's not cyclical.
- First, we need to find all the epimorphisms and count them. There should be as many as there are generating systems.
- Then we need to transfer generating from $F_{2}$ to $Q_{8}$.
- We know that $2$ epimorphisms with same structure differ by automorphism of image, because mutual kernel means that they have the same surjective part.
- Further, we should understand how many automorphisms we have.
- After that just divide the number of epimorphisms by the order of $Q_{8}$ automorphisms group.
And I have the answer, but don't know: it's right or wrong...
The subgroup of the free group in this case is only one. This can be proved in many ways - including through direct verbal description of a suitable sub-group for a list of identities. But the easiest way is probably to see that if $a, b$ - one of the $24$ dyad generators of quaternions, where $a ∈ ${± i, ± j, ± k}$ $, and $b ≠ ± a$, the rule $i↦a, j↦b$ induces an automorphism of $Q_{8}$.
The second method of proof. We know what the ratios of the elements should be performed in the image. For example, the square of any element belonging to the image, belongs to the center. This means that the switches $[x^2, y]$ and $[x, y^2]$ will belong to the kernel of the epimorphism of the free group $F_{2}$ on $Q_{8}$ for any $x, y∈ F_{2}$. Also $x^4$ will belong to the kernel for any $x$. Considering the normal subgroup of the free group generated by the above elements, as well as some others that are obviously lie in the kernel, you can verify that it has an index of $8$, which implies its uniqueness. To verify enough any group word generators result in one of eight types, namely, $a^k b^l [a, b]^m$, wherein $k, l, m∈ ${0, 1}.
Step 6 is incorrect. You have no guarantee that the action of the automorphism group on the set of surjections $F_2 \to Q_8$ is free. What you're trying to compute is the number of orbits under this action (which, for a non-free action, you can't compute just by dividing), which as it turns out is $1$; this is what the paragraph you're quoting proves.