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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?

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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.

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  • $\begingroup$ Why "all non-trivial normal subgroups are of that form" and not all normal subgroups instead (so that I can see how does "+1" come)? $\endgroup$ – DesmondMiles Jun 16 at 23:04
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    $\begingroup$ 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. $\endgroup$ – ThorWittich Jun 16 at 23:07
  • $\begingroup$ 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? $\endgroup$ – DesmondMiles Jun 17 at 0:44
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    $\begingroup$ 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. $\endgroup$ – ThorWittich Jun 17 at 7:29

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