Exercise 14.4.9

Proof. (a) Since $g^2=-I_2$, $g\in \mathrm{GL}(2,{𝔽}_p)$, and since $\mathit{gh}=-\mathit{hg}\ne \mathit{hg}(p>2)$, $g\notin {𝔽}_p^{\text{∗}}{I}_2$. Then Lemma 14.4.3 shows that

if $m,m^{\prime }\in C(g)$, then $m=a{I}_2+\mathit{bg},m^{\prime }=a^{\prime }{I}_2+b^{\prime }g,a,b,a,b{\text{,}}^{\prime }\in F_p$, thus

so $C(g)$ is Abelian.

Moreover, $C(g)$ contains the matrices $a{I}_2\in \mathrm{GL}(2,{𝔽}_p)$, which is equivalent to $a\in {𝔽}_p^{\text{∗}}$.

(b) If $m\in C(g)$, then $m^{2}g=g{m}^2$, thus $g{m}^{-2}=m^{-2}g$ and $(\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2})g=g(\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2})$, so that $\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2}\in C(g)$. This permits us to define

For $m,m^{\prime }\in C(g)$, $\varphi (m{m}^{\prime })=(\det <!--\; FUNCTION\; APPLICATION\; -->(m{m}^{\prime }){(m{m}^{\prime })}^{-2}=(\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2})(\det <!--\; FUNCTION\; APPLICATION\; -->(m^{\prime }){m{}^{\prime }}^{-2})$, thus $\varphi$ is a group homomorphism.

If $m=\lambda I_2,\lambda \in {𝔽}_p^{\text{∗}}$, then $\varphi (m)={\lambda }^2{(\lambda I_2)}^{-2}=I_2$. Therefore ${𝔽}_p^{\text{∗}}{I}_2\subset \ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )$.

Conversely, if $m\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )$, $\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2}=I_2$, thus $m^2=\mu I_2$, where $\mu =\det <!--\; FUNCTION\; APPLICATION\; -->(m)\in \underset{p}{\overset{\text{∗}}{𝔽}}$.

Reasoning by contradiction, suppose that $m\notin {𝔽}_p^{\text{∗}}{I}_2$. Since $m\in C(g)$, Lemma 14.4.3 shows that $m=a{I}_2+\mathit{bg}$, where $b\ne 0$.

Then ${(a{I}_2+\mathit{bg})}^2=\mu I_2$, which gives, using $g^2=-I_2$,

If $a\ne 0$, then $g=-{(2\mathit{ab})}^{-1}(a^2-b^2-\mu )I_2\in {𝔽}_p^{\text{∗}}{I}_2$, which is false. Therefore $a=0$, and $m=\mathit{bg}$. Then

thus $-I_2=g^2=\det <!--\; FUNCTION\; APPLICATION\; -->(g)I^2$, and $\det <!--\; FUNCTION\; APPLICATION\; -->(g)=-1$, which contradicts $\det <!--\; FUNCTION\; APPLICATION\; -->(g)=1(p>2)$. This contradiction shows that $m\in {𝔽}_p^{\text{∗}}{I}_2$. To conclude,

Since $\mathrm{Im}(\varphi )\simeq C(g)∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )=C(g)∕{𝔽}_p^{\text{∗}}{I}_2$,

(c) If $w\in \mathrm{Im}(\varphi )$, then $w=\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2}$ for some $m\in C(g)$. Then

thus

(d) In the proof of part (c) of Proposition 14.4.4, we saw that there is some $Q\in \mathrm{GL}(2,{𝔽}_p)$ such that

Note that

Indeed, if $m\in C(g)$, then $\mathit{mg}=\mathit{gm}$, thus

This proves $Q^{-1}C(g)Q\subset C(g^{\prime })$. Symmetrically, $\mathit{QC}(g^{\prime })Q^{-1}\subset C(g)$, thus $C(g^{\prime })=Q^{-1}C(g)Q.$

Let ${\varphi }^{\prime }:C(g^{\prime })\to C(g^{\prime })$ the group homomorphism defined by ${\varphi }^{\prime }(m^{\prime })=\det <!--\; FUNCTION\; APPLICATION\; -->(m^{\prime }){m{}^{\prime }}^{-2}$. Suppose that we can prove that $\mathrm{Im}({\varphi }^{\prime })=\{w^{\prime }\in C(g^{\prime })\mid \det <!--\; FUNCTION\; APPLICATION\; -->(w^{\prime })=1\}.$ Now, take $w\in C(g)$ such that $\det <!--\; FUNCTION\; APPLICATION\; -->(w)=1$. Since $C(g)=\mathit{QC}(g^{\prime })Q^{-1}$, there is some $w^{\prime }\in C(g^{\prime })$ such that

By our hypothesis, $w^{\prime }\in \mathrm{Im}({\varphi }^{\prime })$, so that $w^{\prime }=\det <!--\; FUNCTION\; APPLICATION\; -->(m^{\prime }){m{}^{\prime }}^2,m^{\prime }\in C(g^{\prime })$.

Put $m=Q{m}^{\prime }{Q}^{-1}$. Then $m\in C(g)$, and

This proves that $\mathrm{Im}(\varphi )=\{w\in C(g)\mid \det <!--\; FUNCTION\; APPLICATION\; -->(w)=1\},$ if we can show first that $\mathrm{Im}({\varphi }^{\prime })=\{w^{\prime }\in C(g^{\prime })\mid \det <!--\; FUNCTION\; APPLICATION\; -->(w^{\prime })=1\}.$ This explain why we may assume now that $g=\left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$.

For all $a,b\in {𝔽}_p$,

By Exercise 5, for all $\lambda \in {𝔽}_p$, there is some pair $(a,b)\in {𝔽}_p^2$ such that $a^2+b^2=\lambda$. If $\lambda \in {𝔽}_p^{\text{∗}}$, then $\det <!--\; FUNCTION\; APPLICATION\; -->(\mathit{aI}+\mathit{bg})=\lambda \ne 0$, thus $a{I}_2+\mathit{bg}\in \mathrm{GL}(2,{𝔽}_p)$. Moreover $a{I}_2+\mathit{bg}\in C(g)$ (this is the easy part of Lemma 14.4.3), thus $\det <!--\; FUNCTION\; APPLICATION\; -->:C(g)\to \underset{p}{\overset{\text{∗}}{𝔽}}$ is a surjective homomorphism.

For any $w\in \mathrm{GL}(2,{𝔽}_p)$,

Thus

Now take any $w=\left(\begin{array}{cc}\hfill r\hfill & \hfill t\hfill \\ \hfill s\hfill & \hfill u\hfill \end{array}\right)\in C(g)$ such that $\det <!--\; FUNCTION\; APPLICATION\; -->(w)=1$.

Then $w=\left(\begin{array}{cc}\hfill r\hfill & \hfill -s\hfill \\ \hfill s\hfill & \hfill r\hfill \end{array}\right),r^2+s^2=1$, and $w^{-1}=\left(\begin{array}{cc}\hfill r\hfill & \hfill s\hfill \\ \hfill -s\hfill & \hfill r\hfill \end{array}\right)$.

Put $m=\left(\begin{array}{cc}\hfill 1+r\hfill & \hfill s\hfill \\ \hfill -s\hfill & \hfill 1+r\hfill \end{array}\right)$ (we found this matrix by trigonometrical analogy). By the previous remark, $m\in C(g)$, and $\det <!--\; FUNCTION\; APPLICATION\; -->(m)={(1+r)}^2+s^2=2(1+r)$.

Then, using $r^2+s^2=1$,

Thus $w=\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2}=\varphi (m)$, where $m=\left(\begin{array}{cc}\hfill 1+r\hfill & \hfill s\hfill \\ \hfill -s\hfill & \hfill 1+r\hfill \end{array}\right)\in C(g)$.

(We have not used with this method the surjectivity of $\det <!--\; FUNCTION\; APPLICATION\; -->:C(g)\to \underset{p}{\overset{\text{∗}}{𝔽}}$.)

We can conclude, using also part (c),

(for $g=\left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$, and by the above remark, for any $g\in \mathrm{GL}(2,{𝔽}_p)$ such that $g^2=-I_2$.)

This equality shows that every element of $C(g)$ of determinant $1$ is of the form $\det <!--\; FUNCTION\; APPLICATION\; -->(m)m^{-2}$ for some $m\in C(g)$. □