Exercise 14.4.10

Proof. (a) In this part, $p\equiv 1(\mathit{mod}4)$. Then $\left(\frac{-1}{p}\right)={(-1)}^{\frac{p-1}{2}}=1$, therefore there exists some element $i\in {𝔽}_p$ such that $i^2=1$.

By Proposition 14.4.4, $H\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ is given, up to conjugacy, by

where

since $s=i,t=0$ is a solution of the equation $s^2+t^2=-1$.

Then $N(H)$ is the normalizer of $H$ in $\mathrm{PGL}(2,{𝔽}_p)$. Write

We verify first that $⟨[a],[b],[c]⟩\subset N(H)$.

Therefore

Since $g,h,k\in H$, we have proved $⟨[a],[b],[c]⟩\subset N(H)$.

Note that $H\subset K=⟨[a],[b],[c]⟩$:

$[k]=[\mathit{ia}]=[a]\in H$, thus $[b][g]{[b]}^{-1}=[k]=[a]$, so that

This proves

Consider the action of $K=⟨[a],[b],[c]⟩\subset N(H)$ on $H\setminus \{[I_2]\}$ by conjugation. This gives a group homomorphism

$\text{∙}$ $\phi$ is surjective: ${\phi }_{[b]}$ is the transposition $(gk)$, and ${\phi }_{[c]}$ is the transposition $(hk)$. Since $S_3$ is generated by any pair of distinct transpositions, $S(\{[g],[h],[k]\}=⟨(gk),(hk)⟩$. This proves that $\phi$ is surjective. $\text{∙}$ Since $H$ is Abelian, the elements of $H$ fix $[g],[h],[k]$ by conjugation, therefore $H\subset \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$. To prove the converse, we note that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\subset C(H)=H$, because every element of $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ fixes $[I_2],[g],[h],[k]$, thus is in $C(H)$, which is $H$ (Proposition 14.4.4).

Therefore $S_3\simeq \mathrm{Im}(\phi )\simeq K∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=K∕H$. This gives

Since $⟨[a],[b],[c]⟩\subset N(H)$, and $|⟨[a],[b],[c]⟩|=24=|N(H)|$, we obtain

If we replace our $g$ and $h$ by any pair $g,h$ satisfying the conditions of Proposition 14.4.4, then $N(H)$ is a conjugate subgroup of $K=⟨[a],[b],[c]⟩$ by part (c) of this Proposition. (b) In this part, $p\equiv 3(\mathit{mod}4)$. There is no element $i$ of order 4, but by Proposition 14.1.4, there exist $s,t\in {𝔽}_p$ such that

where

and

Write

Put $K=⟨[g],[h],[c],[d]⟩$. We want to show that $N(H)=K=⟨[g],[h],[c],[d]⟩.$

We know that $H\subset N(H)$, so $[g],[h]\in N(H)$. It remains to verify that $[c],[d]\in N(H)$.

Since $g,h,k\in H$, it follows that

$K=⟨[g],[h],[c],[d]⟩\subset N(H)$.

As in part (a), consider the homomorphism

Then ${\varphi }_{[c]}$ is the transposition $(hk)$, and ${\varphi }_{[d]}$ is the transposition $(gh)$. Since $S_3$ is generated by any pair of two distinct transpositions, $S([g],[h],[k])=⟨(hk),(gh)$, thus $\varphi$ is surjective. Since $H\subset K$, as in part (a), $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=H$.

Therefore $S_3\simeq \mathrm{Im}(\phi )\simeq K∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=K∕H$. This gives

Since $⟨[g],[h],[c],[d]⟩\subset N(H)$, and $|⟨[g],[h],[c],[d]⟩|=24=|N(H)|$, we obtain

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