Exercise 14.4.8

Proof. First, we try to understand the first sentence, and why this is related to the proof of Proposition 14.4.4.

Note that every permutation of $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ fixing $(0,0)$ is an outer automorphism (the Cayley tables are identical if we exchange the two elements of any pair of nonzero elements). Since $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ is Abelian, there is no non trivial inner automorphism. Thus $\mathrm{Aut}(H)\simeq S_3$.

In the proof of Proposition 14.4.4, if we write $G=N(H)=N_{\mathrm{PGL}(2,{𝔽}_p)}(H)$, then we proved that

is surjective (onto). Note that $\phi$ is well defined because $H$ is a normal subgroup of $G$, so $\mathit{gh}{g}^{-1}\subset H$ for all $g\in G$ and for all $h\in H$.

By hypothesis, $\phi$ is surjective. As in the text, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=C_G(H)=H$, thus

This proves $|G|=|H|\times |S_3|=4\times 6=24$. (a) Let $N$ be the number of $3$-Sylow subgroups of $G$. By the third Sylow Theorem,

Therefore $N=1$ or $N=4$.

If $N=1$, then there is only one $3$-Sylow subgroup, say $K$. Since $\mathit{gK}{g}^{-1}$ is also a $3$-Sylow subgroup, $\mathit{gK}{g}^{-1}=K$, so $K$ is normal in $G$.

Since $|K|=3$ and $|H|=4$, then $|H\cap K|$ divides $4$ and $3$, thus $H\cap K=\{e\}$.

By Exercise 14.3.7, knowing that $H,K$ are normal in $G$, and that $H\cap K=\{e\}$, then $\mathit{HK}$ is a normal subgroup of $G$, and $\mathit{HK}\simeq H\times K$. Moreover $H$ and $K$ are Abelian subgroups, thus $\mathit{HK}\simeq H\times K$ is abelian. This shows that $K\subset C_G(H)$, thus $H=C_G(H)\supset \mathit{HK}⊋H$. This is a contradiction.

We have proved that $G$ has four $3$-Sylow subgroups. (b) Let $H_1,H_2,H_3,H_4$ be the $3$-Sylow subgroup of $G$. For all $g\in G$, $g{H}_i{g}^{-1}$ is a $3$-Sylow of $G$, thus $G$ acts on the set $\{H_1,H_2,H_3,H_4\}$.

By the second Sylow Theorem, all $3$-Sylow subgroups are conjugate, thus $G$ acts transitively on $\{H_1,H_2,H_3,H_4\}$, thus the orbit of $H_1$ is

By definition of the normalizer, the isotropy group of $H_1$ is $G_{H_1}=N_G(H_1)$. The Fundamental Theorem of Group Actions gives

thus $|N_G(H_1)|=|G|∕4=6$. (c) Write $E=\{H_1,H_2,H_3,H_4\}$. Consider the homomorphism

$(M=H_i,i=1,2,3,4).$

Reasoning by contradiction, suppose that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$ contains an element $h$ of order $3$.

For all $g\in G$, $g\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$ if and only if ${\psi }_g$ is the identity of $S(E)$, thus

Therefore $\{e,h,h^2\}$ is a subgroup of $N_G(H_1)$. But $H_1\subset N_G(H_1)$ is another subgroup of $N_G(H_1)$ of order $3$.

Since $N_G(H_1)=6$, $N_G(H_1)$ has a unique subgroup of order $3$: the number $n$ of $3$-Sylow subgroups of $N_G(H_1)$ satisfies $n\mid 6,n\equiv 1(\mathit{mod}3)$, thus $n=1$. Therefore $H_1=\{e,h,h^2\}$. Reasoning with $H_2\subset N_G(H_2)$, we obtain similarly $H_2=\{e,h,h^2\}$. Thus $H_1=H_2$. This is a contradiction since the $3$-Sylow subgroups $H_1,H_2,H_3,H_4$ are distinct. Therefore $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$ cannot contain an element of order $3$. (d) By part (b), $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$ is a subgroup of the group $N_G(H_1)$, whose order is $6$. By part (c), $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$ has no element of order 3, hence the order of $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$ is not $3$ or $6$, otherwise, by Cauchy Theorem, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$ would contain an element of order $3$. Thus the order of $\ker <!--\; FUNCTION\; APPLICATION\; -->(\varphi )$ is $1$ or $2$. This implies that

If we choose some numbering $\gamma$ of $E$, where $\gamma :\{1,2,3,4\}\to E$ is given by $\gamma (i)=H_i$, then $\hat{\gamma }:S(E)\to S_4$ defined by $\hat{\gamma }(\sigma )={\gamma }^{-1}\circ \sigma \circ \gamma$ is an isomorphism. Thus $\varphi =\hat{\gamma }\circ \psi$, $\varphi :G\to S_4$, is a group homomorphism, such that $|\mathrm{Im}(\varphi )|=|\mathrm{Im}(\psi )|=12$ or $24$. Since $S_4$ has only one subgroup of order $12$, we obtain that

We have proved $\mathrm{Im}(\varphi )\supset A_4$.

If $\mathrm{Im}(\varphi )=S_4$, then $\varphi :G\to S_4$ is surjective, and since $|G|=|S_4|$, $\varphi$ is bijective, thus $\varphi$ is a group isomorphism and $G\simeq S_4$.

It follows that if $\varphi$ is not an isomorphism, then $\mathrm{Im}(\varphi )=A_4$, and $|\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )|=2$. Then $K=\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ is a normal subgroup of $G$ of order $2$. (e) It remains to show that $G$ cannot contain a normal subgroup $K$ of order $2$. Reasoning by contradiction, suppose that there exists such a subgroup $K$. Write $K=\{e,a\},a\ne e$.

Since $K$ is normal in $G$, for every $g\in G$, $\mathit{ga}{g}^{-1}\in K=\{e,a\}$, and $\mathit{ga}{g}^{-1}\ne e$, otherwise $a=e$. Thus $\mathit{ga}{g}^{-1}=a$.

This means that $a$ is in the center of $G$, a fortiori, $a\in C_G(H)$. Since $C_G(H)=H$, it follows that

Write $H=\{e,a,b,c\}$.

Consider anew the homomorphism $\phi :G\to \mathrm{Aut}(H)\simeq S_3$ described in the introduction. By hypothesis, $\phi$ is surjective. But for all $g\in G$, ${\phi }_g(a)=\mathit{ga}{g}^{-1}=a$, thus any automorphism $\chi$ of $G$ such that $\chi (a)\ne a$ , for instance the automorphism defined by $\chi (e)=e,\chi (a)=b,\chi (b)=c,\chi (c)=a$, is not in $\mathrm{Im}(\varphi )$, in contradiction with the surjectivity of $\phi$.

This proves that $G$ has no normal subgroup of order 2. By part (d), it follows that $S(E)\simeq S_4$. □