Exercise 13.2.7

Proof.

(a)

Let $G$ be the symmetry group of $u$.

$\text{∙}$

If $\sigma \in G$, then $\sigma \cdot u=u$, therefore $\sigma \cdot h=\sigma \cdot u^2=u^2=h$. By Lemma 13.2.4, $\sigma \in \mathrm{AGL}(1,{𝔽}_5)$, so $G\subset \mathrm{AGL}(1,{𝔽}_5)$. $G\ne \mathrm{AGL}(1,{𝔽}_5)$, otherwise $(1243)\in G$, but $(1243)\cdot u=-u\ne u$ (see (13.2.B)). Therefore $G⊊\mathrm{AGL}(1,{𝔽}_5)$.

Moreover $(12345)\cdot u=u$, so $⟨(12345)⟩\subset G$ and $G$ is transitive. By Theorem 13.2.2,

$$G\subset \mathrm{AGL}(1,{𝔽}_5)\cap A_5.$$

$\text{∙}$

If $\chi \in \mathrm{AGL}(1,{𝔽}_5)\cap A_5$, by Exercise 1 part (b), $$\chi ={(12345)}^k{[(14)(23)]}^l,k,l\in \mathbb{N}.$$

$(12345)\cdot u=u$ and $(1243)\cdot u=-u$, therefore $(14)(23)\cdot u={(1243)}^2\cdot u=u$. Thus $\chi \in G$.

$$G=\mathrm{AGL}(1,{𝔽}_5)\cap A_5.$$

(b)

In Exercise 4, we verified that for $u,v\in S,u\ne v$, with $$S=\{e,(123),(234),(345),(145),(125)\}\subset A_5,$$

then $u{v}^{-1}\notin \mathrm{AGL}(1,{𝔽}_5)$, a fortiori $u{v}^{-1}\notin \mathrm{AGL}(1,{𝔽}_5)\cap A_5$.

Moreover the index $(A_5:\mathrm{AGL}(1,{𝔽}_5)\cap A_5)=60∕10=6$, so $S$ is a complete system of coset representatives of $\mathrm{AGL}(1,{𝔽}_5)\cap A_5$ in $A_5$.