9 final

Problem 1.

For subgroup $H,K$ of a finite group $G$, prove that $G=HK$ iff $|G|=|H||K|/|H\cap K|$.

Proof. We have proved it in class. And this is a proof from Shuan yang Hu.

Problem 2.

(Recall Hall theorem: Let $G$ be a finite solvable group and let $\pi (G)$ be the set of prime divisors of $|G|$. Then for any subset $\pi \subset \pi (G)$, there exists a Hall $\pi$-subgroup $G_{\pi }$ of $G$.)

Prove that, for any Sylow $p$-subgroup and Hall $p^{\prime }$-subgroup $G_{p^{\prime }}$ of $G$, we have that $G=G_p{G}_{p^{\prime }}$.

Proof. If $H$ is a Hall $\pi$-subgroup of $G$, then the order and index of $H$ are relatively prime, that is, $(|H|,|[G:H]|)=1$. Suppose $|G|=p^e{p}_1^{e_1}\cdots p_k^{e_k}$, where $p_i$ and $p$ are distinct primes. By Sylow theorem, $|G_p|=p^e$. And the order of Hall $p^{\prime }$-subgroup $G_{p^{\prime }}$ is $p_1^{e_1}\cdots p_k^{e_k}$. Since $G_p{G}_{p^{\prime }}$ is a subset of $G$ and $|G_p{G}_{p^{\prime }}|=|G_p||G_{p^{\prime }}|/|G_p\cap G_{p^{\prime }}|=|G|$, $G=G_p{G}_{p^{\prime }}$.

Problem 3.

Let $K<G$, and let $\Omega =[G:K]$. Prove that, for any subgroup $H<G$, $G=HK$ iff $H$ is transitive on $\Omega$.

• Let $G=A_n$ and $K=A_{n-2}$. Prove that $G$ has a subgroup $H={\mathrm{PSL}}_d(q)$ with $n=\frac{q^d-1}{q-1}$, such that $G=HK$.
• Let $G=S_n$ and $K=S_{n-3}$, where $n=q+1$ with $q$ being a prime-power. Then $G$ has a subgroup $H={\mathrm{PGL}}_2(q)$ such that $G=HK$.

Proof. By Frattini’s argument. Let $\mathcal{P}$ be the set of $1$-space of ${\mathbb{F}}_p^d$, then $n=|\mathcal{P}|$. Since ${\mathrm{PSL}}_d(q)$ is $2$-transitive on $\mathcal{P}$ and ${\mathrm{PGL}}_2(q)$ is $3$-transitive on $\mathcal{P}$, then $A_n={\mathrm{PSL}}_d(q)A_{n-2}$ and $S_n={\mathrm{PGL}}_2(q)S_{n-3}$. Note that ${\mathrm{PSL}}_d(q)$ is simple, it is a subgroup of $A_n$. But ${\mathrm{PGL}}_2(q)$ may not a subgroup of ${\mathrm{A}}_{\mathrm{n}}$, so it cannot always be written as $A_n={\mathrm{PGL}}_2(q)A_{n-2}$.

Problem 4.

Let $G=HK$ and $N⊲G$. Let $\overline{G}=G/N$ and $\overline{H},\overline{K}$ the images of $H,K$ under $G\to G/N$. Then $\overline{G}=\overline{H}\overline{K}$.

Give an example to show $\overline{G}=\overline{H}\overline{K}$ may be trivial factorization.

Proof. Recall that $G/N$ is the group with elements $\{Ng:g\in G\}$ and multiplication $(N{g}_1)(N{g}_2)=N(g_1{g}_2)$ for $g_1,g_2\in G$. Select $Ng\in G/N$. As $G=HK$ there is $h\in H,k\in K$ such that $g=hk$, and we obtain $Ng=(Nh)(Nk)\in \overline{H}\overline{K}$. Thus $\overline{G}\subset \overline{H}\overline{K}$, and $\overline{H}\overline{K}\subset \overline{G}$ holds trivially. So $\overline{G}=\overline{H}\overline{K}$.

For a trivial factorization, consider $N\ge H$. For example $G=⟨a⟩\times ⟨b⟩=C_2\times C_2$ where $H=⟨a⟩,K=⟨b⟩$. Let $N=⟨a⟩⊲G$. Then $\overline{G}=\overline{K}$ and $\overline{H}$ is trivial.

Rmk. Here is an nontrivial example. Let $T={\mathrm{PSL}}_2(p)$ such that $12\nmid p-1$ and $12\nmid p+1$. Let $H=⟨a⟩:⟨b⟩=S_3$. Let $g\in C_T(b)\backslash H$ such that $|g|=2$. Let $S=g^H=\{g,g^a,g^{a^2}\}$. Then $H$ is $2$-transitive on $S$. Let $M=⟨H,g⟩\le T$.

Claim that $M=T$. Otherwise, assume $M\le L{<}_{\max }T$, then $L\in \{Z_p:Z_{(p-1)/2},D_{p-1},D_{p+1},A_4,S_4,A_5\}.$Since $L>S_3$ and $12||L|$, $L$ does not exist. So $M=T$.

Claim that $N=T$. Let $N=⟨S⟩$. Since $g,a$ and $b$ normalize $N$, $1\ne N⊲⟨g,a,b⟩=M=T$. By $T$ simple, $N=T$.

Let $\Gamma =\mathrm{Cay}(T,S)$. Then $T$ is connected and $\tilde{H}<\mathrm{Aut}(\Gamma )$. And $G:=\hat{T}:\tilde{H}=\hat{T}\times \check{H}$. The group $G$ is $2$-arc-transitive on $\Gamma$. And $T=G/\check{H}$ is $2$-arc-transitive on ${\Gamma }_{\check{H}}$. Then $\hat{T}\cong \overline{G}=\overline{\hat{T}}\overline{\tilde{H}}=\overline{\hat{T}}$. Furthermore, we obtain that ${\Gamma }_{\check{H}}$ is non-Cayley, which I don’t know how to prove. For example, when $p=5$, ${\Gamma }_{\check{H}}$ is Petersen.

Problem 5.

Let $G=HK$ with $H,K$ both abelian. Prove that $G$ is a meta-abelian group. Is this true that if $H,K$ are metacyclic then $G=HK$ metacyclic? Why?

Def. If the commutator subgroup $G^{\prime }$ is abelian, we say $G$ is metabelian.

Proof. i) First, we show that $[H,K]$ is normal. For any $hk{h}^{-1}{k}^{-1}\in [H,K]$, we have $(hk{h}^{-1}{k}^{-1}{)}^{h_0}=h{k}^{h_0}{h}^{-1}(k^{h_0}{)}^{-1}$. Suppose $k^{h_0}=k^{\prime }{h}^{\prime }$, then $(hk{h}^{-1}{k}^{-1}{)}^{h_0}=h{k}^{\prime }{h}^{-1}{k}^{\prime -1}$. Similarly, consider $h^{k_0}=h^{\prime \prime }{k}^{\prime \prime }$ and then $(hk{h}^{-1}{k}^{-1}{)}^{k_0}=h^{\prime \prime }k{h}^{\prime \prime -1}{k}^{-1}$.

Then claim that $[G,G]=[H,K]$. For any $g_1,g_2\in G$, suppose $g_1=h_1{k}_1$ and $g_2=k_2{h}_2$. Then $g_1{g}_2{g}_1^{-1}{g}_2^{-1}=h_1{k}_2{k}_1{h}_2{k}_1^{-1}{h}_2^{-1}{h}_1^{-1}{k}_2^{-1}=[h_1,k_2][k_1,h_2{]}^{h_1^{-1}{k}_2^{-1}}\in [H,K]$. Therefore, $[G,G]=[H,K]$.

Finally, prove $[G,G]=[H,K]$ is abelian. For any $h\in H,k\in K$, $[h,k{]}^{\tilde{h}\tilde{k}}=[h,k{]}^{\tilde{k}\tilde{h}}$. Then $[h,k]$ and $[\tilde{h},\tilde{k}]$ are commute.

ii) Product of cyclic groups may not metacyclic. Consider $G=D_6\times D_6=Z_6{Z}_6$. Note that $G$ is not metacyclic. Otherwise, assume $A⊲G$ is cyclic and $G/A$ is cyclic. Then $A\ge G^{\prime }=C_3\times C_3$ is not cyclic, contradiction.

Rmk. Similarly, we can prove that nilpotent*nilpotent=solvable.

Problem 6.

Prove that $G={\mathrm{PSL}}_2(p)\times {\mathrm{PSL}}_2(p)$ is a product of two meta-cyclic groups iff $p\equiv 3\mathrm{mod}4$.

Problem 7.

Let $G=HK$ where $H$ is a simple group and $|K|\le 100$. Prove that, there are only finitely many possibilities such that $H$ is not a normal subgroup of $G$.

Proof. Since $H$ is simple and not a normal subgroup of $G$, $H$ is core-free in $G$. Hence, the right multiplication action of $G$ on $[G:H]$ is faithful, i.e. $G\le \mathrm{Sym}([G:H])$. Recall that $G=HK$ implies $K$ is a transitive subgroup of $\mathrm{Sym}([G:H])$. Then by $|K|<100$, one has the lengths of orbits of $K$ are less than $100$. Therefore, $G\le \mathrm{Sym}([G:H])\le S_{100}$, that is, there are only finitely many possibilities.

Problem 8.

Prove that most arc-transitive cubic graphs are non-Cayley graphs.

(Recall Tutte’s theorem: If $\Gamma$ is a $G$-arc-transitive cubic graph, then the vertex stabilizer $G_v$ is a subgroup of $S_4\times S_2$.)

Proof. We cannot finish this problem completely, but we can prove something related to it.

Suppose $\Gamma =\mathrm{Cay}(G,S)$ is an arc-transitive cubic graph, where $G$ is a finite group and $S$ is a $3$-subset of $G$. Then for any vertex $\alpha \in V\Gamma$, $X:=\mathrm{Aut}\Gamma =G{X}_{\alpha }$. Let $C={\mathrm{Core}}_X(G)$. Then there are three cases.

i) If $C=1$, $G$ is core-free in $X$. Thus $X$ acting on $\Omega :=[X:G]$ is faithful and so $X\le \mathrm{Sym}(\Omega )\le S_{48}$ by Tutte’s theorem. So there are only finitely many possibilities of $G$.

ii) If $C=G$, then $\Gamma$ is a normal arc-transitive graph.

iii) If $1<C<G$, we consider the quotient graph ${\Gamma }_C$. Since $\mathrm{val}({\Gamma }_C)|\mathrm{val}(\Gamma )$, we have $\mathrm{val}(\Gamma )=1$ or $3$.

If $\mathrm{val}{\Gamma }_C=1$, then ${\Gamma }_C$ is a single edge. Then $\Gamma$ is a bipartite graph, that is, any $C$-orbit is a independent set, because $C$-orbits is a block system and $\Gamma$ is arc-transitive. This graph is called binormal Cayley graph.

If $\mathrm{val}{\Gamma }_C=3$, then $\Gamma$ is a cover of ${\Gamma }_C$ and ${\Gamma }_C$ is a arc-transitive cubic graph. Since $G/C$ acting on $V{\Gamma }_C$ is regular, ${\Gamma }_C$ is a Cayley graph and $G/C$ is core-free. By i) we know there are finitely many possibilities of $G/C$.

Problem 9.

For suitable integer $n$, express the complete graph $K_n$ as a normal edge-transitive Cayley graph.

Lemma. $\Gamma$ is a normal edge-transitive Cayley graph iff $\mathrm{Aut}(G,S)$ is transitive on $\{\{g,g^{-1}\}:g\in S\}$.

Proof. For any $g_1,g_2\in S$, suppose $g_1$ and $g_2$ are not conjugate in $\mathrm{Aut}(G,S)$. There is $\hat{g}h\in N$ such that $\{1,g_1{\}}^{\hat{g}h}=\{1,g_2\}$ where $\hat{g}\in \hat{G}$ and $h\in \mathrm{Aut}(G,S)$. Then $g^h\ne 1$ and $(g{g}_1{)}^h=1$, $g=g_1^{-1}$ and $(g_1^{-1}{)}^h=g_2$. Thus $g_1$ and $g_2^{-1}$ are conjugate in $\mathrm{Aut}(G,S)$.

Proof. Suppose $K_n=\mathrm{Cay}(G,G\backslash \{1\})$ is normal edge-transitive. Then $\mathrm{Aut}(G,G\backslash \{1\})$ is transitive on $D:=\{\{g,g^{-1}\}:g\in G,g\ne 1\}$ and so all non-trivial elements of $G$ have the same order. So there is a prime $p$ such that $|g|=p$ for all $g\ne 1$. Since $G$ is a $p$-group, $Z(G)$ is non-trivial. Since $Z(G)$ is a characteristic subgroup and $\mathrm{Aut}(G)$ is transitive on $D$, $Z(G)=G$. Thus $G=Z_p^d$ and $n=p^d$ for some positive integer $d$.

When $G=Z_p^d$, $\mathrm{Aut}(G,G\backslash \{1\})=\mathrm{GL}(d,p)$ is transitive on $G\backslash \{1\}$. Therefore, $K_n=\mathrm{Cay}(G,G\backslash \{1\})$ is a normal edge-transitive Cayley graph.

Problem 10.

For suitable integers $n$, express the complete multi-partite graph $K_{n,\cdots ,n}$ as a normal edge-transitive Cayley graph.

Problem 11.

Construct $2$-arc-transitive Cayley graphs $\Gamma$ such that $\Gamma$ have normal quotient graphs which are non-Cayley graphs.

Does there exist a normal $3$-arc-transitive Cayley graph?

Proof. i) Let $G=A_5$, acting on $\{1,2,3,4,5\}$, and let $H=⟨(123),(23)(45)⟩\cong S_3$ and $g=(24)(35)$. Let $S=g^H=\{(24)(35),(34)(15),(14)(25)\}$ and $\Gamma =\mathrm{Cay}(G,S)$.

Claim that $⟨S⟩=G$ and hence $\Gamma$ is connected. Since $(24)(35{)}^{(34)(15)}=(23)(45)$ and $(24)(35)(34)(15)(14)(25)=(12345)$, we have $(23)(45),(12345)\in ⟨S⟩$. And $A_5=⟨(23)(45),(12345)⟩$ yields $⟨S⟩=G$.

Claim that $\Gamma$ is $2$-arc-transitive. Since $H$ is a maximal subgroup of $G$, we have $\mathrm{Inn}(G,S)=\tilde{H}$. Let $X=\hat{G}:\mathrm{Inn}(G,S)$. Then $X=\hat{G}\times \check{H}\cong A_5\times S_3$. Since $\mathrm{Inn}(G,S)$ is $2$-transitive on $S$, $\Gamma$ is $(X,2)$-arc-transitive. Thus $\Gamma$ is a $2$-arc-transitive graph.

Moreover, let $N=\check{H}$, the quotient graph ${\Gamma }_N$ is the Petersen graph, which is non-Cayley.

ii) The answer is negative. Rmk. I think it is enough to consider these two arcs: $1-s_1-s_1{s}_2-s_1{s}_2{s}_3$ and $1-s_1-s_1{s}_2-s_1{s}_2{s}_1$. Since $\mathrm{Aut}\Gamma =\hat{G}:\mathrm{Aut}(G,S)$, there does not exist $\sigma$ mapping one $3$-arc to another.

Problem 12.

Prove that, for any integer $k\ge 3$, there are at most finitely many graphs of valency $k$ which are $4$-arc-transitive Cayley graphs.

(Recall that, if $\Gamma$ is $(G,4)$-arc-transitive graph of valency $k\ge 3$ which is not $(G,5)$-arc transitive, then $G_v⊳q^2:{\mathrm{SL}}_2(q)$ and $G_v^{\Gamma (v)}⊳{\mathrm{PSL}}_2(q)$, with $k=q+1$.)

Proof. Use three lemmas:

• 1992, Praeger (theorem 1)
• lemma 2, $X_v=A_7$ or $S_7$
• ${\mathrm{Core}}_{X}G$ acting on $V$ is transitive (the last part)

Problem 13.

Let $\Gamma$ be a self-complementary undirected graph of order $n$ with a complementing isomorphism $\sigma$. Prove that the order of $\sigma$ is divisible by $4$.

Proof. Since $\Gamma$ is self-complementary, we have $n\equiv 0$ or $1\mathrm{mod}4$. For the nontrivial cases, we can assume $n\ge 4$. If $|\sigma |$ iss divisible by some odd $m$, then ${\sigma }^m$ is also a complementing isomorphism of $\Gamma$, as $\sigma$ interchanges $\Gamma$ and $\overline{\Gamma }$. Replacing $\sigma$ by ${\sigma }^m$, we can assume $|\sigma |=2^e$ for some integer $e$. By the definition of a complementing isomorphism, $\sigma$ does not fix any pair of vertices in $\Gamma$. Clearly $e\ge 1$. For any $2$ distinct vertices $u,v$, let $w=u^{\sigma },z=v^{\sigma }$, then $\{u,v{\}}^{\sigma }=\{w,z\}\ne \{u,v\}$. WLOG assume $w\ne u$. If $|\sigma |=2$, $\{u,w{\}}^{\sigma }=\{w,u\}$, which is a contradiction. Thus we have $|\sigma |$ is divisible by $4$.

Rmk. I think my proof is easier than this. WLOG $|\sigma |=2^e$ for some positive integer $e$. If $e$ is $1$, then there is a vertex $u$ such that $u^{\sigma }=v,v^{\sigma }=u$. Then $\{u,v{\}}^{\sigma }=\{u,v\}$, whch is impossible.

Problem 14.

Prove that edge-transitive self-complementary graphs are normal Cayley graphs are normal Cayley graphs.

Problem 15.

Prove that, if a group $G$ has an automorphism $\sigma$ such that ${\sigma }^2$ fixes no non-identity element of $G$, and each orbit of $⟨\sigma ⟩$ on $G\backslash \{1_G\}$ is of even length, then there exists a self-complementary Cayley graph $\Gamma =\mathrm{Cay}(G,S)$ such that $\sigma$ is a complementing isomorphism of $\Gamma$.

Proof. Without the italic section, the problem has errors. We will prove it by three parts:

• with the condition of “orbit with even length”, we can construct a SC Cayley graph
• if $\sigma$ is a complementing isomorphism of SC Cayley graph, orbits of $\sigma$ on $G\backslash \{1_G\}$ is of even length
• a counter example such that $\sigma \in \mathrm{Aut}(G)$, ${\sigma }^2$ fixes no non-identity elements, but SC $\mathrm{Cay}(G,S)$ does not exist.

Lemma. $\sigma$ is a complementing isomorphism of Cayley graph $\Gamma =\mathrm{Cay}(G,S)$ iff $S^{\sigma }=G\backslash (S\cup \{1_G\}$.

Proof. If $\sigma$ is a complementing isomorphism, then $(1,g)$ is either an edge of $\Gamma (1_G)$ or an edge of $\overline{\Gamma }(1_G)$. So $G=S\bigsqcup S^{\sigma }\bigsqcup \{1_G\}$.

Conversely, assume $S^{\sigma }=G\backslash (S\cup \{1_G\})$, where $\sigma \in \mathrm{Aut}(G)$ and $S$ is a subset of $G$. Suppose $\Gamma =\mathrm{Cay}(G,S)$. For any $(x,y)\in E\Gamma$, we have $(x,y{)}^{\sigma }\notin E\Gamma$. Also $(z,w)\notin E\Gamma$ yields $(z,w{)}^{\sigma }\in E\Gamma$. Thus $\Gamma$ is self-complementary.

i) Assume each orbit of $⟨\sigma ⟩$ on $G\backslash \{1_G\}$ is of even length. Let $G=1_G\bigsqcup g_1^{⟨\sigma ⟩}\bigsqcup \cdots \bigsqcup g_k^{⟨\sigma ⟩}$ be an orbit decomposition of $⟨\sigma ⟩$ on $G$. Since each orbit of $⟨\sigma ⟩$ is of even length, that is, the order of $\sigma$ is even, we have $g_i^{⟨\sigma ⟩}=g_i^{⟨{\sigma }^2⟩}\bigsqcup g_i^{\sigma ⟨{\sigma }^2⟩}$ and $\sigma$ permutes $g_i^{⟨{\sigma }^2⟩}$ and $g_i^{\sigma ⟨{\sigma }^2⟩}$.

Let $S=g_1^{⟨{\sigma }^2⟩}\bigsqcup \cdots \bigsqcup g_k^{⟨{\sigma }^2⟩}$, then $S^{\sigma }=G\backslash (S\cup \{1_G\})$. And hence by lemma $\sigma$ is a complementing isomorphism of $\mathrm{Cay}(G,S)$.

ii) On the other hand, assume $\sigma \in \mathrm{Aut}(G)$ and ${\sigma }^2$ fixes no elements of $G\backslash \{1_G\}$, and there is a SC $\Gamma =\mathrm{Cay}(G,S)$ with self-complementing isomorphism $\sigma$. Assume $g\in S$ and the orbit of $g$ under $⟨\sigma ⟩$ is $\{g,g^{\sigma },\cdots ,g^{{\sigma }^k}\}$. Then $\{g^{\sigma },g^{{\sigma }^3},\cdots \}\in G\backslash (S\cup \{1_G\})$ and $\{g,g^{{\sigma }^2},\cdots \}\in S$, and so $|g^{⟨\sigma ⟩}|$ is even. If $1_G\ne g\notin S$, then $S^{\sigma }=G\backslash (S\cup \{1_G\})$ yields $g\in S^{\sigma }$. With the same argument $|g^{⟨\sigma ⟩}|$ is even. Therefore, all orbits of $⟨\sigma ⟩$ on $G\backslash \{1_G\}$ are of even length.

iii) Counter example: $G=⟨x⟩=⟨(1234567)⟩\cong C_7$, $\sigma :x\to x^2$. Then ${\sigma }^2$ fixs no non-identity elements but $\mathrm{Cay}(G,S)$ is not SC because $7\not\equiv 0$ or $1\mathrm{mod}4$.