Exercise 13.2.1

Proof.

(a)

Let $r={\gamma }_{1,1}:{𝔽}_5\to {𝔽}_5,x\mapsto x+1$ and $s={\gamma }_{2,0}:{𝔽}_5\to {𝔽}_5,x\mapsto 2x$, corresponding to the permutations $\rho =(12345)$ and $\sigma =(1243)$.

Since 2 is a generator of ${𝔽}_5^{\text{∗}}$ ( $2^2\equiv -1\mathit{mod}5$), every $a\in {𝔽}_p^{\text{∗}}$ is of the form $a=2^k,k\in \mathbb{N}$, so every $f\in \mathrm{AGL}(1,{𝔽}_5)$, defined by $x\mapsto \mathit{ax}+b,a=2^k\in {𝔽}_p^{\text{∗}},b\in {𝔽}_p$ is equal to $f=r^b\circ s^k$. Therefore $\mathrm{AGL}(1,{𝔽}_5)=⟨r,s⟩$, and the corresponding subgroup $G$ of $S_5$, isomorphic to $\mathrm{AGL}(1,{𝔽}_5)$, is generated by $\rho =(12345)$ and $\sigma =(1243)$.

(b)

By part (a), every permutation $\chi$ of $\mathrm{AGL}(1,{𝔽}_5)$ is of the form $\chi ={\rho }^b{\sigma }^k,0\le b\le 4,0\le k\le 3$. Since $\rho \in A_5$ and $\sigma \in S_5\setminus A_5$, $\chi \in A_5$ if and only if $k$ is even. Moreover, since ${\sigma }^4=e$, for each integer $l$, ${\sigma }^{2l}=e$ or ${\sigma }^{2l}={\sigma }^2$, so $$\begin{array}{llll}\hfill \mathrm{AGL}(1,{𝔽}_5)\cap A_5& =\{{\rho }^k|0\le k\le 4\}\cup \{{\rho }^k{\sigma }^2|0\le k\le 4\}& \hfill & \\ \hfill & =\{e,\rho ,{\rho }^2,{\rho }^3,{\rho }^4,{\sigma }^2,\rho {\sigma }^2,{\rho }^2{\sigma }^2,{\rho }^3{\sigma }^2,{\rho }^4{\sigma }^2\}.& \hfill & \end{array}$$

Thus

$$\mathrm{AGL}(1,{𝔽}_5)\cap A_5=⟨\rho ,{\sigma }^2⟩=⟨(12345),(14)(23)⟩.$$

(c)

For every $x\in {𝔽}_p$, $(s^2\circ r)(x)=4(x+1)$, and $(r^{-1}\circ s^2)(x)=4x-1=4x+4=4(x+1)$, so $s^2\circ r=r^{-1}\circ s^2$ and ${\sigma }^2\rho ={\rho }^{-1}{\sigma }^2$.

Write ${\sigma }^{\prime }={\sigma }^2$. Since $\mathrm{AGL}(1,{𝔽}_5)\cap A_5=⟨\rho ,{\sigma }^{\prime }⟩$, the relations

$${\rho }^5=e,{\sigma {}^{\prime }}^2=e,{\sigma }^{\prime }\rho ={\rho }^{-1}{\sigma }^{\prime }$$

characterize the dihedral group $D_{10}$.

(d)

Let $H⊋⟨(12345)⟩$ be a subgroup of $\mathrm{AGL}(1,{𝔽}_5)$. By part (a), $H$ contains an element ${\rho }^b{\sigma }^k$, with $k\in \{1,2,3\}$.

Since $\rho \in H$, ${\sigma }^k={\rho }^{-b}({\rho }^b{\sigma }^k)\in H$.

If $k=1$, then $\sigma \in H$, and if $k=3$, then ${\sigma }^3={\sigma }^{-1}\in H$. In both cases, $\sigma \in H$. Since $\mathrm{AGL}(1,{𝔽}_5)$ is generated by $\rho =(12345)$ and $\sigma =(1243)$, then $H=\mathrm{AGL}(1,{𝔽}_5)$.

It remains the case where $H$ contains ${\sigma }^2$ and doesn’t contain $\sigma$. Then $H\supset ⟨\rho ,{\sigma }^2⟩$. No element of the form ${\rho }^b{\sigma }^{2k+1}$ is in $H$, otherwise $\sigma \in H$, so

$$H=⟨\rho ,{\sigma }^2⟩=\mathrm{AGL}(1,{𝔽}_5)\cap A_5.$$

Thus the only subgroups of $\mathrm{AGL}(1,{𝔽}_5)$ containing $⟨(12345)⟩$ are

$$⟨(12345)⟩,\mathrm{AGL}(1,{𝔽}_5)\cap A_5,\mathrm{AGL}(1,{𝔽}_5).$$