Exercise 14.2.16

Proof. (a) Recall that the group of invertible elements of $\mathbb{Z}∕p^2\mathbb{Z}$ is ${(\mathbb{Z}∕p^2\mathbb{Z})}^{\text{∗}}$, where

As in section 6.4, if $a,b\in \mathbb{Z}∕p^2\mathbb{Z}$, we define ${\gamma }_{a,b}:\mathbb{Z}∕p^2\mathbb{Z}\to \mathbb{Z}∕p^2\mathbb{Z}$ by ${\gamma }_{a,b}(x)=\mathit{ax}+b$. Note that ${\gamma }_{a,b}$ is bijective if and only if $a\in {(\mathbb{Z}∕p^2\mathbb{Z})}^{\text{∗}}$. We define

We note that ${\gamma }_{a,b}={\gamma }_{c,d}$ implies ${\gamma }_{a,b}(0)={\gamma }_{c,d}(0)$, thus $b=d$, and then ${\gamma }_{a,b}(1)={\gamma }_{c,d}(1)$ gives $a=c$:

Therefore the map

is bijective, which proves $|G|=\varphi (p^2)p^2=(p^2-p)p^2=p^3(p-1)$.

As in section 6.4 (and Exercise 6.4.1), we obtain, if $a,c\in {(\mathbb{Z}∕p^2\mathbb{Z})}^{\text{∗}}$,

since $\mathit{ac}\in {(\mathbb{Z}∕p^2\mathbb{Z})}^{\text{∗}}$. Moreover, ${\gamma }_{a,b}^{-1}={\gamma }_{a^{-1},-a^{-1}b}\in G$, so that $G$ is a subgroup of $S(\mathbb{Z}∕p^2\mathbb{Z})\simeq S_{p^2}$.

The map

is well-defined, and is a surjective homomorphism by definition of $G$. Moreover, the kernel of $\phi$ is the subgroup

Therefore $\mathrm{AGL}(1,\mathbb{Z}∕p^2\mathbb{Z})∕T\simeq {(\mathbb{Z}∕p^2\mathbb{Z})}^{\text{∗}}$. Since $T$ and ${(\mathbb{Z}∕p^2\mathbb{Z})}^{\text{∗}}$ are Abelian, this proves that $G=\mathrm{AGL}(1,\mathbb{Z}∕p^2\mathbb{Z})$ is solvable.

To prove that $G$ acts transitively on $\mathbb{Z}∕p^2\mathbb{Z}$, for all $\alpha ,\beta \in \mathbb{Z}∕p^2\mathbb{Z}$, we must find ${\gamma }_{a,b}\in G$ such that ${\gamma }_{a,b}(\alpha )=\beta$.

Write $\alpha ={[u]}_{p^2},\beta ={[v]}_{p^2},a={[A]}_{p^2},b={[B]}_{p^2}$, where $u,v,A,B\in \mathbb{Z}$, and $\gcd <!--\; FUNCTION\; APPLICATION\; -->(p,A)=1$. Then

If $\alpha \in {(\mathbb{Z}∕p^2\mathbb{Z})}^{\text{∗}}$, $b=0$ and $a={\alpha }^{-1}\beta$ give a solution ${\gamma }_{a,b}={\gamma }_{{\alpha }^{-1}\beta ,0}$.

Otherwise, $p\mid u$. If $p^2\nmid u$, then $u=\mathit{kp},k\in \mathbb{Z},p\nmid k$. We can take $B=v-p$, so that

which has a solution $A\in \mathbb{Z}$ since $p\nmid A$.

Finally, if $p^2\mid u$, then $\alpha =0$, and $a=1,b=\beta$ gives a solution ${\gamma }_{1,\beta }$.

Thus $G$, viewed as a subgroup of $S_{p^2}$, is transitive. (b) Consider the subgroup $H$ of $G$ defined by

where we write, for every integer $a$, $\bar{a}={[a]}_{p^2}$ the class of $a$ modulo $p^2$.

Then the cosets $R_i=\bar{i}+H,0\le i\le p-1$, partition $G$:

Note that $R_i={\gamma }_{1,i}(R_0)$ for all $i$ (here ${\gamma }_{1,i}$ refers to ${\gamma }_{\bar{1},\bar{i}}$, and $i+H$ to $\bar{i}+H=R_i$).

We prove

Indeed, if $\alpha =\overline{\mathit{kp}}\in R_0$, then ${\gamma }_{a,b}(\overline{\mathit{kp}})=a\overline{\mathit{kp}}+b\in b+H=R_b$, thus ${\gamma }_{a,b}(R_0)\subset R_b$. Moreover the cosets $R_0,R_b$ have same cardinality, so that ${\gamma }_{a,b}(R_0)=R_b$.

Using this result, for all index $i$, since $R_i=i+R_0={\gamma }_{1,i}(R_0)$,

Therefore $G$ permutes the blocks $R_0,\dots ,R_{p-1}$.

This proves that $G=\mathrm{AGL}(1,\mathbb{Z}∕p^2\mathbb{Z})$ is imprimitive. □