# Normal subgroups of invertible affine transformations $\pmod p$

Fix a prime $$p$$. Find the number of normal subgroups of the group $$G$$ of invertible affine maps $$x \to bx + c$$, $$b \neq 0$$ on $$\mathbb{Z}/p\mathbb{Z}$$.

It is clear that the group has cardinality $$p(p-1)$$ and is generated by $$x \to x + 1$$ and $$x \to gx$$ where $$g$$ is a primitive root mod p and it's easily verifiable that the former gives a normal subgroup.

Update: What if we want the number of all normal subgroups $$H$$ such that the quotient $$G/H$$ is Abelian?

Let me denote that group by $$\text{Aff}(\mathbb{F}_p)$$ and its elements by $$f_{b,c}$$. Consider the homomorphism $$\varphi \colon \text{Aff}(\mathbb{F}_p) \rightarrow \mathbb{F}_p^{\times}$$, $$f_{b,c} \mapsto b$$. Since $$\mathbb{F}_p^{\times}$$ is abelian, all its subgroups $$H \subset \mathbb{F}_p^{\times}$$ are normal subgroups. Therefore we have that $$\varphi^{-1}(H)$$ is a normal subgroup of $$\text{Aff}(\mathbb{F}_p)$$ for every subgroup $$H \subset \mathbb{F}_p^{\times}$$. Actually, all non-trivial normal subgroups of $$\text{Aff}(\mathbb{F}_p)$$ are of that form. This means you need to count the number of subgroups of the cyclic group $$\mathbb{F}_p^{\times}$$. Thus we have $$1 + \sigma(p-1)$$ normal subgroups, where $$\sigma$$ is the divisor function.
• Well... The trivial group cannot be realized as a preimage under this map since the $c$ is still arbitrary. Therefore I counted the number of subgroups arising that way and added one for the trivial subgroup. – ThorWittich Jun 16 at 23:07
• Yep, thanks! Do you have an idea how does the answer change if we additionally insist that for each normal subgroup $H$ the quotient $Aff(\mathbb{F}_p)/H$ is Abelian? – DesmondMiles Jun 17 at 0:44
• Yes, all of them yield abelian quotients except for the trivial one. That is because the commutator subgroup of $\text{Aff}(\mathbb{F}_p)$ coincides with the kernel of the map $\varphi$ such that every non-trivial normal subgroup contains the commutator subgroup. This means we have $\sigma(p-1)$ normal subgroups yielding abelian quotients. – ThorWittich Jun 17 at 7:29