Exercise 14.4.12

Proof. We recall the Second and Third Isomorphism Theorems (see for instance [14] Rose, A course on Group Theory, p. 77-79).

Second Isomorphism Theorem. If $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then $N$ is a normal subgroup of $\mathit{HN}\subset G$, and

$$H∕H\cap N\simeq \mathit{HN}∕N.$$

Third Isomorphism Theorem. If $K,H$ are normal subgroups of $G$ such that $K\subset H$, then $H∕K$ is a normal subgroup of $G∕K$, and

$$G∕H\simeq (G∕K)∕(H∕K).$$

(a) By (8) in Exercise 14.3.3, the map

$$\{\begin{array}{ccc}\hfill {𝔽}_{p^2}^{\text{∗}}\times \mathrm{Gal}({𝔽}_{p^2}∕{𝔽}_p)\hfill & \hfill \to \hfill & \hfill {(M_1)}_0=\{{\gamma }_{a,\sigma ,v}\in M_1\mid v=0\}\hfill \\ \hfill (a,\sigma )\hfill & \hfill \mapsto \hfill & \hfill {\gamma }_{a,\sigma ,0}\hfill \end{array}$$

is bijective, thus

$$|{(M_1)}_0|=2(p^2-1).$$

Note that ${(M_1)}_0\cap \mathrm{AGL}(1,{𝔽}_{p^2})=\{{\gamma }_{a,0}\in \mathrm{AGL}(1,{𝔽}_{p^2})\mid a\in {𝔽}_{p^2}^{\text{∗}}\}$. Indeed, if ${\gamma }_{a,\sigma ,0}\in \mathrm{AGL}(1,{𝔽}_{p^2})$, where $\sigma \in \mathrm{Gal}({𝔽}_{p^2}∕{𝔽}_p)=\{e,F\}$. If $\sigma =F$ is the Frobenius isomorphism, there are some $c,d\in {𝔽}_{p^2},c\ne 0$ such that ${\gamma }_{a,\sigma ,0}={\gamma }_{c,d}$, thus, for all $u\in {𝔽}_{p^2}$,

$$a{u}^p=\mathit{cu}+d.$$

For $u=0$, we obtain $d=0$, and for $u=1$, we obtain $a=c$, thus $u^p=u$, which implies $u\in {𝔽}_p$. Since ${𝔽}_p⊊{𝔽}_{p^2}$, this is a contradiction, therefore $\sigma =e$, and ${\gamma }_{a,\sigma ,0}={\gamma }_{a,0}\in \mathrm{GL}(1,{𝔽}_{p^2})\subset \mathrm{AGL}(1,{𝔽}_{p^2})$.

We have proved

$$K={(M_1)}_0\cap \mathrm{AGL}(1,{𝔽}_{p^2})=\{{\gamma }_{a,0}\in \mathrm{AGL}(1,{𝔽}_{p^2})\mid a\in {𝔽}_{p^2}^{\text{∗}}\}\simeq {𝔽}_{p^2}^{\text{∗}},$$

therefore the subgroup $K$ of ${(M_1)}_0$ is Abelian, with order $|{𝔽}_{p^2}^{\text{∗}}|=p^2-1$, thus $K$ is an Abelian subgroup of index 2 in ${(M_1)}_0$.

Similarly, ${(M_2)}_0$ is the subgroup of $M_2$ generated by $\mathrm{GL}(1,{𝔽}_p)\times \mathrm{GL}(1,{𝔽}_p)$ and $B=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$, thus

$$\begin{array}{llll}\hfill {(M_2)}_0& =\left\{\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right)\mid a,b\in {𝔽}_p^{\text{∗}}\right\}\cup \left\{\left(\begin{array}{cc}\hfill 0\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill 0\hfill \end{array}\right)\mid a,b\in {𝔽}_p^{\text{∗}}\right\}& \hfill & \\ \hfill & =L\cup (B\cdot L),& \hfill & \end{array}$$

where $L$ is the group of diagonal matrices in $\mathrm{GL}(2,{𝔽}_p)$, with order $|L|={(p-1)}^2$, and $L\simeq {𝔽}_p^{\text{∗}}\times {𝔽}_p^{\text{∗}}$ is Abelian, of index $2$ in ${(M_2)}_0$, where $|{(M_2)}_0|=2{(p-1)}^2$.

Reasoning by contradiction, suppose that ${(M_3)}_0\subset {(M_1)}_0$. The idea is to prove that this implies the existence of an Abelian subgroup of index $2$ in $S_4$, which is false.

Consider the Abelian subgroup $K$ of index $2$ in ${(M_1)}_0$, whose existence is proved above, and the following diagram, where all arrows are inclusions.

The subgroup $K$ has index $2$ in ${(M_1)}_0$, thus is a normal subgroup of ${(M_1)}_0$. Hence $K{(M_3)}_0$ is a subgroup of ${(M_1)}_0$, and

$$K\subset K{(M_3)}_0\subset {(M_1)}_0.$$

Since

$$2=({(M_1)}_0:K)=({(M_1)}_0:K{(M_3)}_0)(K{(M_3)}_0:K),$$

we obtain $(K{(M_3)}_0:K)=1$ or $2$. Moreover, if $(K{(M_3)}_0:K)=1$, then $K{(M_3)}_0=K$ and ${(M_3)}_0\subset K$. But this is impossible since $K$ is Abelian, but ${(M_3)}_0$ is not ( ${\gamma }_{A,0}\mapsto [A]$ maps ${(M_3)}_0$ on $N(H)\simeq S_4$, which is not Abelian). Hence $(K{(M_3)}_0:K)=2$, thus $K{(M_3)}_0={(M_1)}_0$. Then the Second Isomorphism Theorem shows that

$${(M_3)}_0∕K\cap {(M_3)}_0={(M_1)}_0∕K=2.$$

If we write $M=K\cap {(M_3)}_0$, then $M$ is an Abelian subgroup of ${(M_3)}_0$ of index $2$ in ${(M_3)}_0$.

Note that $K={(M_1)}_0\cap \mathrm{AGL}(1,{𝔽}_{p^2})$, thus

$$M=K\cap {(M_3)}_0=({(M_1)}_0\cap \mathrm{AGL}(1,{𝔽}_{p^2}))\cap {(M_3)}_0={(M_3)}_0\cap \mathrm{AGL}(1,{𝔽}_{p^2}).$$

Moreover, if we identify ${\gamma }_{A,0}$ with $A\in \mathrm{GL}(2,{𝔽}_p)$, then $\{{\gamma }_{\lambda I_2,0}\mid \lambda \in {𝔽}_p^{\text{∗}}\}\simeq {𝔽}_p^{\text{∗}}{I}_2$, and for all $\lambda \in {𝔽}_{p^2}^{\text{∗}}$, ${\gamma }_{\lambda I_2,0}\in {(M_3)}_0\cap \mathrm{AGL}(1,{𝔽}_{p^2})$, therefore ${𝔽}_p^{\text{∗}}{I}_2\subset M$.

$M$ is an Abelian subgroup of ${(M_3)}_0$ of index $2$ in ${(M_3)}_0$ containing ${𝔽}_p^{\text{∗}}{I}_2$. It remains to prove that this is impossible.

Consider the group homomorphism (where $A$ is identified with ${\gamma }_{A,0}$).

$$\phi \{\begin{array}{ccc}\hfill {(M_3)}_0\hfill & \hfill \text{↦}\hfill & N(H)\subset \mathrm{PGL}(2,{𝔽}_p)\hfill \\ \hfill A\hfill & \hfill \text{↦}\hfill & [A]\hfill \end{array}$$

Since ${(M_3)}_0=\{A\in \mathrm{GL}(2,{𝔽}_p)\mid [A]\in N(H)\}$, $\phi$ is surjective, and $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )={𝔽}_p^{\text{∗}}{I}_2\subset M$.

In the following diagram, where $P=\phi (M)$,

the horizontal arrows are inclusions, and the vertical arrows are restrictions of $\phi$.

The restriction ${\phi }_1\{\begin{array}{ccc}\hfill M\hfill & \hfill \text{↦}\hfill & P=\phi (M)\hfill \\ \hfill g\hfill & \hfill \text{↦}\hfill & \phi (g)\hfill \end{array}$ is surjective, and $\ker <!--\; FUNCTION\; APPLICATION\; -->({\phi }_1)=\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\cap M=\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$.

By the First Isomorphism Theorem,

$$M∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\simeq P=\phi (M),{(M_3)}_0∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\simeq N(H)=\phi ({(M_3)}_0).$$

Then the Third Isomorphism Theorem gives

$$N(H)∕P\simeq ({(M_3)}_0∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi ))∕(M∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi ))\simeq {(M_3)}_0∕M\simeq \{-1,1\}.$$

Since $\phi (M)=P$, $P$ is Abelian, and is a subgroup of index $2$ in $N(H)\simeq S_4$. Hence $S_4$ would contain an Abelian subgroup of index 2, but this is impossible, since the only subgroup of index 2 in $S_4$ is $A_4$, which is not Abelian.

We have proved ${(M_3)}_0\not\subset {(M_1)}_0$. Now, if $M_3\subset M_1$, then

$$\{g\in M_3\mid g\cdot 0=0\}\subset \{g\in M_1\mid g\cdot 0=0\},$$

so ${(M_3)}_0\subset {(M_1)}_0$, which is false. Therefore

$$M_3\not\subset M_1.$$

Since $M_2$ has also an Abelian subgroup of index 2 which contains ${𝔽}_p^{\text{∗}}{I}_2$, we prove similarly that

$$M_2\not\subset M_1.$$

(b) By Exercise 14.3.26, for $p=3$, we obtain

$$\begin{array}{llll}\hfill |\mathrm{AGL}(2,{𝔽}_3)|& =p^3(p-1)(p^2-1)& \hfill & \\ \hfill & =27\times 2\times 8=432,& \hfill & \end{array}$$

and by Exercise 14.4.11,

$$\begin{array}{llll}\hfill |M_3|& =24p^2(p-1)& \hfill & \\ \hfill & =24\times 9\times 2=432.& \hfill & \end{array}$$

Since $M_3\subset \mathrm{AGL}(2,{𝔽}_3)$ by definition, and $|M_3|=|\mathrm{AGL}(2,{𝔽}_3)|$, we obtain

$$M_3=\mathrm{AGL}(2,{𝔽}_3)(p=3).$$

(c) We recall (see Cox, p. 453, and Ex. 16) that ${(M_1)}_0$ is the group $\mathrm{\Gamma L}(1,{𝔽}_{p^2})$ consisting of semi-linear maps ${\gamma }_{\alpha ,\sigma }:{𝔽}_{p^2}\to {𝔽}_{p^2},\alpha \in {𝔽}_{p^2},\sigma \in \mathrm{Gal}({𝔽}_{p^2}∕{𝔽}_p)=\{e,F\}$, linear over ${𝔽}_p$, defined by

$${\gamma }_{\alpha ,\sigma }(u)=\mathit{\alpha \sigma }(u),u\in {𝔽}_{p^2},$$

so that ${(M_1)}_0\subset \mathrm{GL}(2,{𝔽}_p),$ and we can consider the group ${(M_1)}_0∕{𝔽}_p^{\text{∗}}{I}_2$.

Now, let $c$ be a generator of ${𝔽}_{p^2}^{\text{∗}}$, and consider $f={\gamma }_{c,e}\in {(M_1)}_0$ defined by $f(u)=\mathit{cu}$ for $u\in {𝔽}_{p^2}$, and $M\in \mathrm{GL}(2,{𝔽}_p)$ the matrix representing the ${𝔽}_p$-linear map $f$.

Then for all $u\in {𝔽}_{p^2}$, $f^{p+1}(u)=c^{p+1}u=\mathit{hu}$, where $h^{p-1}={(c^{p+1})}^{p-1}=1$, which proves that $h=c^{p+1}\in {𝔽}_p^{\text{∗}}$. Therefore the matrix representing $f^{p+1}$ is $\left(\begin{array}{cc}\hfill h\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill h\hfill \end{array}\right)$, so that ${[M]}^{p+1}=[I_2].$ The order of $[f]$ in ${(M_1)}_0∕{𝔽}_p^{\text{∗}}{I}_2$ divides $p+1$.

Conversely, if ${[M]}^k=[I_2]$ for some integer $k>0$, then $M^k=\lambda I_2$ for some $\lambda \in {𝔽}_p^{\text{∗}}$, thus

$$\forall <!--\; FUNCTION\; APPLICATION\; -->u\in {𝔽}_{p^2},f^k(u)=c^{k}u=\mathit{\lambda u}.$$

Therefore $c^k=\lambda \in {𝔽}_p^{\text{∗}}$, which implies $c^{k(p-1)}=1$. Since the order of $c$ is $o(c)=p^2-1$, we obtain $p^2-1\mid k(p-1)$, thus $p+1\mid k$. This proves that the order of $[f]$ (or [M]) in ${(M_1)}_0∕{𝔽}_p^{\text{∗}}{I}_2$ is $p+1$.

To conclude, ${(M_1)}_0∕{𝔽}_p^{\text{∗}}{I}_2$ contains elements of order $p+1$.

It remains to prove that $M_1\not\subset M_3$ when $p>3$.

If $M_1\subset M_3$, then ${𝔽}_p^{\text{∗}}{I}_2\subset {(M_1)}_0\subset {(M_3)}_0$, therefore

$${(M_1)}_0∕{𝔽}_p^{\text{∗}}{I}_2\subset {(M_3)}_0∕{𝔽}_p^{\text{∗}}{I}_2.$$

We proved in part (a) that ${(M_3)}_0∕{𝔽}_p^{\text{∗}}{I}_2\simeq N(H)\simeq S_4$.

The first group ${(M_1)}_0∕{𝔽}_p^{\text{∗}}{I}_2$ contains an element of order $p+1$, which is also in ${(M_3)}_0∕{𝔽}_p^{\text{∗}}{I}_2\simeq S_4$, and this implies that

$$p+1\mid 24=|{(M_3)}_0∕{𝔽}_p^{\text{∗}}{I}_2|,$$

which gives $p=5,7,11,23$ (if $p>3$). But in these cases we obtain in $S_4$ some elements of order $6,8,12$ or $24$. But all elements in $S_4$ have orders $1,2,3$ or $4$. This is a contradiction.

To conclude,

$$M_1\not\subset M_3\text{ if }p>3.$$

Example. Take $p=11$. Since $x^2-4d-9$ is irreducible over ${𝔽}_{11}$, we use the basis $\mathcal{}B=(1,c)$ of ${𝔽}_{p^2}$ over ${𝔽}_p$, where $c$ is a generator of ${𝔽}_{11}^{\text{∗}}$ such that $c^2=4c+9$. Then the matrix of $f$ defined by $f(u)=\mathit{cu}$ in the basis $\mathcal{}B$ is

$$A=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 9\hfill \\ \hfill 1\hfill & \hfill 4\hfill \end{array}\right).$$

The multiplicative order of $A$ is 120, but with Sage, we obtain

$$A^{12}=\left(\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right),$$

(where $2=c^{12}$), so that ${[A]}^{12}=[I_2]$.

The matrix $S$ corresponding to the Frobenius isomorphism $\sigma =F$ is, using $\sigma (c)=c^{11}=10c+4$,

$$S=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 10\hfill \end{array}\right).$$

Then $S^2=I_2$, and we can build the whole group ${(M_1)}_0$ by

$${(M_1)}_0=\{A^k\mid 0\le k<120\}\cup \{A^{k}S\mid 0\le k<120\}.$$

The only elements of order $p+1$ in ${(M_1)}_0∕{𝔽}_p^{\text{∗}}{I}_2$ are in the $[A^k]$, and not in the $[A^{k}S]$. (d) By Propositions 14.4.2 and 14.4.5, we know that

$$\begin{array}{llll}\hfill |M_2|& =2p^2{(p-1)}^2,& \hfill & \\ \hfill |M_3|& =24p^2(p-1).& \hfill & \end{array}$$

If $M_2\subset M_3$, by Lagrange Theorem, $|M_2|=2p^2{(p-1)}^2$ divides $|M_3|=24p^2(p-1)$, therefore

$$p-1\mid 12.$$

The only prime solution $p$ with $p>5$ are $p=7,p=13$.

If $p=13$, then $|M_2|=|M_3|=48672=2^5{3}^{2}13^2$, and $M_2\subset M_3$, thus $M_2=M_3$. But this is impossible, since, by part (a), ${(M_2)}_0$ has a subgroup of index 2 containing ${𝔽}_p^{\text{∗}}{I}_2$, but not ${(M_3)}_0$.

It remains to prove that $M_2\not\subset M_3$ if $p=7$. Then $|M_2|=3528=2^3{3}^2{7}^2=3528,|M_3|=2^4{3}^2{7}^2=7056$, thus $(M_3:M_2)=2$, and $M_2$ would be a normal subgroup of index $2$ in $M_3$.

Consider the surjective homomorphism

$$\phi \{\begin{array}{ccc}\hfill M_3\hfill & \hfill \text{↦}\hfill & N(H)\subset \mathrm{PGL}(2,{𝔽}_p)\hfill \\ \hfill {\gamma }_{A,v}\hfill & \hfill \text{↦}\hfill & [A].\hfill \end{array}$$

Note that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{{\gamma }_{\lambda I_2,v}\mid \lambda \in {𝔽}_p^{\text{∗}},v\in {𝔽}_p^2\}\subset M_2$, since for all $u=(\alpha ,\beta )\in {𝔽}_p^2$, and $v=(a,b)$

$${\gamma }_{\lambda I_2,v}=\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \lambda \hfill \end{array}\right)\left(\begin{array}{c}\hfill \alpha \hfill \\ \hfill \beta \hfill \end{array}\right)+\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \mathit{\lambda \alpha }+a\hfill \\ \hfill \mathit{\lambda \beta }+b\hfill \end{array}\right),$$

so that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\subset \mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)\subset M_2$.

Write $N=\mathrm{AGL}(1,{𝔽}_p)\times \mathrm{AGL}(1,{𝔽}_p)$, and $M=\phi (N)$. In the following diagram,

the horizontal arrows are inclusions, and the vertical arrows are restrictions of $\phi$, whose kernels are $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, because $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\subset N$. Therefore

$$N(H)∕L\simeq (M_3∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi ))∕(M_2∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi ))\simeq M_3∕M_2\simeq \{-1,1\}.$$

This proves that $L$ is a subgroup of index 2 in $N(H)\simeq S_4$. But $S_4$ has a unique subgroup of order 2, which is $A_4$. Thus $L\simeq A_4$.

Similarly, $N$ has index 2 in $M_2$, thus $M$ has index 2 in $L\simeq A_4$. This is impossible, since $A_4$ has no subgroup of index $2$.

To conclude: If $p>5$,

$$M_2\not\subset M_3.$$

(e) If $p=5$, $|M_2|=2p^2{(p-1)}^2=800,|M_3|=24p^2(p-1)=2400$.

Here $p\equiv 1(\mathit{mod}4)$, and $i=2$ is such that $i^2=-1$. By definition of $H$, $H=⟨[g],[h]⟩$, where

$$g=\left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right),h=\left(\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -2\hfill \end{array}\right)$$

(by Proposition 14.4.4, other choices of $g,h$ as in part (a) give conjugate subgroups). Thus

$$H=\left\{[I_2],\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right],\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right],\left[\begin{array}{cc}\hfill -2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]\right\}\simeq C_2\times C_2.$$

We have seen in part (a) that

$$\phi \{\begin{array}{ccc}\hfill {(M_3)}_0\hfill & \hfill \text{↦}\hfill & N(H)\subset \mathrm{PGL}(2,{𝔽}_p)\hfill \\ \hfill A\hfill & \hfill \text{↦}\hfill & [A]\hfill \end{array}$$

satisfies $\phi ({(M_3)}_0)=N(H)$, and that

$$\begin{array}{llll}\hfill {(M_2)}_0& =\left\{\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right)\mid a,b\in {𝔽}_5^{\text{∗}}\right\}\cup \left\{\left(\begin{array}{cc}\hfill 0\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill 0\hfill \end{array}\right)\mid a,b\in {𝔽}_5^{\text{∗}}\right\}& \hfill & \\ \hfill & =L\cup (B\cdot L).& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill \phi ({(M_2)}_0)& =\left\{\left[\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b\hfill \end{array}\right]\mid a,b\in {𝔽}_5^{\text{∗}}\right\}\cup \left\{\left[\begin{array}{cc}\hfill 0\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill 0\hfill \end{array}\right]\mid a,b\in {𝔽}_5^{\text{∗}}\right\}& \hfill & \\ \hfill & =\phi (L)\cup ([B]\cdot \phi (L)).& \hfill & \end{array}$$

Since $[I_2],\left[\begin{array}{cc}\hfill -2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]\in \phi (L)$, and $\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right],\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]\in [B]\cdot \phi (L)$, we obtain

$$H\subset \phi ({(M_2)}_0),$$

where $\phi ({(M_2)}_0)\simeq {(M_2)}_0∕{𝔽}_p^{\text{∗}}{I}_2$ has order $|{(M_2)}_0|∕|{𝔽}_p^{\text{∗}}|=32∕4=8$.

Since $H$ is a subgroup of index $2$ in $\phi ({(M_2)}_0)$, $H$ is a normal subgroup of $\phi ({(M_2)}_0)$, so that $\phi ({(M_2)}_0)$ normalizes $H$:

$$\phi ({(M_2)}_0)\subset N(H)=\phi ({(M_3)}_0).$$

If $A\in {(M_2)}_0$, then $[A]=\phi (A)\in \phi ({(M_2)}_0)\subset \phi ({(M_3)}_0)=N(H)$, so that $A\in {(M_3)}_0$.

$${(M_2)}_0\subset {(M_3)}_0.$$

Then we can prove $M_2\subset M_3$. Let $\gamma ={\gamma }_{A,v}$ be any element in $M_2\subset \mathrm{AGL}(2,{𝔽}_p)$. Then $\gamma ={\gamma }_{I_2,v}{\gamma }_{A,0}$, where ${\gamma }_{A,0}\in {(M_2)}_0\subset {(M_3)}_0$, and ${\gamma }_{I_2,v}\in M_3$. Therefore $\gamma \in M_3$.

If $p=5$,

$$M_2\subset M_3.$$

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