Burnside's Transfer Theorem

transfer

Let $H<G$ be a subgroup and $\Omega =[G:H]=\{H{x}_1,\cdots ,H{x}_n\}$. Define $\hat{g}:Hx\mapsto Hxg$ and ${\tau }_g\in \mathrm{Sym}(n)$ be the permutation induced from $\hat{g}$. Since $H{x}_{i}g=H{x}_{i^{\tau (g)}}$, we have that $x_{i}g{x}_{i^{\tau (g)}}^{-1}\in H$.

Define $h_i(g)=x_{i}g{x}_{i^{\tau (g)}}^{-1}$, and define

$$\begin{array}{rl}{V}_{G\to H}:G& \to H/H^{\prime },\\ g& \mapsto ({\Pi }_{i=1}^n{h}_i(g))H^{\prime }\end{array}$$

as a transfer from $G$ to $H/H^{\prime }$.

Observation.

• $V_{G\to H}$ is a homomorphism from $G$ to $H/H^{\prime }$;
• $V_{G\to H}$ does not depend on the choices of $x_1,\cdots ,x_n$;
• Writing $\hat{g}$ in cycle form $\left(H{x}_1,H{x}_{1}g,\cdots ,H{x}_1{g}^{l_1-1}\right)\left(H{x}_2,H{x}_{2}g,\cdots ,H{x}_2{g}^{l_1-2}\right)\cdots \left(H{x}_t,\cdots ,H{x}_t{g}^{l_t-1}\right)$, then there is $V_{G\to H}(g)={\Pi }_{i=1}^n\left(x_i{g}^{l_i}{x}_i^{-1}\right)H$.

p-nilpotent

A group $G$ is called $p$-nilpotent if $G=NP$, where $N⊲G$ and $P$ is a Sylow $p$-subgroup, and $N\cap P=1$.

Burnside's transfer theorem

Let $G$ be a finite group, and let $P$ be a Sylow $p$-subgroup. If $N_G(P)=C_G(P)$, then $G$ is $p$-nilpotent.

\begin{proof} Since $N_G(P)=C_G(P)⩾P$, then $P$ is abelian and $P^{\prime }=1$.

Since $V_{G\to P}(g)={\Pi }_{i=1}^t{x}_i{g}^{l_i}{x}_i^{-1}={\Pi }_{i=1}^t{g}^{l_1+\cdots +l_t}=g^{|G/P|}$, $V_{G\to P}$ is surjective. So $V_{G\to P}:G\to P$ has kernel $K$ and $G/K\cong P$. So $K⊲G$ such that $G=KP$ and so $G$ is $p$-nilpotent. \end{proof}

Corollary

Theorem

Let $G$ be finite of which Sylow subgroups are all cyclic. Let $p$ be the smallest prime divisor of $G$ and let $P$ be a Sylow $p$-subgroup of $G$. Then:

• $G$ is $p$-nilpotent.
• $G$ is solvable.

\begin{proof} Since $P$ is cyclic and $p$ is the smallest prime divisor, we have that $N_G(P)/C_G(P)\lesssim \mathrm{Aut}(P)$ and $N_G(P)/C_G(P)=1$. So $G$ has a normal subgroup $K$ with $|G/K|=|P|$. Then $K$ is $q$-nilpotent where $q>p$ is the smallest prime divisor of $K$. Repeat this procedure and we get $G⊳K⊳K_1\cdots ⊳K_m$. Note that $K_m$ is a Sylow subgroup and $K_i/K_{i+1}$ is solvable. So $G$ is solvable. \end{proof}

Corollary

If a Sylow $2$-group of $G$ is cyclic, then $G$ is solvable.

\begin{proof} By ^sezt3q, $G$ is $2$-nilpotent and so there exists $N⊲G$ such that $G=NP$ and $|N|$ is odd. By Feit–Thompson theorem, $G$ is solvable. \end{proof}

Proposition

Let $G$ be a finite nonabelian simple group, and let $p$ be the smallest prime divisor of $|G|$. Then $|G|$ is divisible by $p^3$ or $12$.

\begin{proof} Suppose $|G|$ is not divisible by $p^3$, then $p^2||G|$. Otherwise, a Sylow $p$-subgroup $P=C_p$ and $G$ is $P$-nilpotent, contradiction by $G$ simple. Thus $P$ is cyclic or $P\cong C_p\times C_p$. The first case is impossible. Hence $P\cong C_p\times C_p$ and $N_G(P)\ne C_G(P)$. Further $N_G(P)/C_G(P)\lesssim \mathrm{Aut}(P)={\mathrm{GL}}_2(p)$ and $|{\mathrm{GL}}_2(p)|=p(p-1{)}^2(p+1)$. Since $p$ is the smallest prime divisor and $G$ is nonsolvable, then $p=2$ and $p+1=3$. It yields that $12||G|$.
\end{proof}

Application

Proposition

Let $|G|=p^2{q}^n$ with $p<q$ primes. Then $G$ is solvable.

\begin{proof} Note that $n_q\equiv 1(\mathrm{mod}q)$ and $n_q\mid p^2$, we have $n_q=p^2$. Let $P$ be a Sylow $p$-subgroup of $G$. If $N_G(P)/C_G(P)=1$, then $G$ is $p$-nilpotent and so $G$ is solvable. Otherwise $N_G(P)/C_G(P)⩽\mathrm{GL}(2,p)$ where $|\mathrm{GL}(2,p)|=(p-1{)}^{2}p(p+1)$. Since $p^2\mid |C_G(P)|$, there is $|N_G(P)/C_G(P)|=p+1$ and so $p=2$, $q=3$. Hence $n_3=4$ and so for a Sylow $3$-group $Q$, $G/Q_G\lesssim S_4$. As $Q_G$ is a $q$-group and is solvable, $G$ is solvable. \end{proof}

Proposition

Group of order $2^3\cdot 3^3\cdot 5=1080$ is not simple.

\begin{proof} Assume that $G$ is simple. Since $6!<1080$, $G$ does not have subgroup of index $⩽6$. Hence $n_5\in \{36,216\}$. If $n_5=216$, $[G:N_G(P)]=216$ and so $N_G(P)=C_G(P)=P$. It deduces that $G$ is $5$-nilpotent, contradiction. Thus $n_5=36$ and $|N_G(P)|=30$. Since $N_G(P)\ne C_G(P)$ and $N_G(P)/C_G(P)\lesssim C_4$, we have $|C_G(P)|=15$ and $C_G(P)$ is cyclic.

Let $H⩽C_G(P)$ be the group of order $3$. We claim that $N_G(P)<N_G(H)$.

• Since $H\mathrm{char}{C}_G(P)$ and $C_G(P)⊲N_G(P)$, there is $H⊲N_G(P)$.
• Consider $N_G(H)$. There exists a Sylow $3$-subgroup $Q⩾H$. If $H⩽Z(Q)$, then $N_G(H)⩾Q⩾27$. Otherwise $H\cap Z(Q)=\{1\}$ and $N_G(H)⩾⟨H,Z(Q)⟩⩾3^2$. Thus $3^2\mid N_G(H)$.
• As $3^2\nmid N_G(P)$, $N_G(P)<N_G(H)$.

Since $[N_G(H):N_{N_G(H)}(P)]=[N_G(H):N_G(P)]>1$, the number of Sylow $5$-subgroups of $N_G(H)$ is not equal to $1$. Since $n_5\ne 216$, the number of Sylow $5$-subgroups of $N_G(H)$ is $6$ or $36$.

• If it equals to $6$, then $|N_G(H)|=6\ast 30=180$ and $[G:N_G(H)]=6$, which is impossible.
• If it equals to $36$, then $N_G(H)=G$ and $H⊲G$, contradicting with $G$ simple.

Now we finish the proof. \end{proof}

Proposition

Group of order $180$ is not simple.

\begin{proof} Since $180=2^2\ast 5\ast 9$, we have $n_5\in \{6,36\}$. If $n_5=36$, then $|G:N_G(P)|=36$ and $N_G(P)=5$, where $P$ is a Sylow $5$-subgroup. It deduces that $N_G(P)=C_G(P)=P$ and so $G$ is $5$-nilpotent. Hence $n_5=6$ and $|N_G(P)|=30$. Since $N_G(P)/C_G(P)\lesssim C_4$, we have $|C_G(P)|=15$ and $C_G(P)$ is cyclic. It yields that $G$ has an element of order $15$.

Since $n_5=[G:N_G(P)]=6$, $G\lesssim S_6$ does not have an element of order $15$. Contradiction. \end{proof}

Proposition

Group of order $2^2\ast 3\ast 5\ast 7=420$ is not simple.

\begin{proof} Assume $G$ is simple. By Sylow’s theorem, $n_7=15$. Then $N_G(P)/C_G(P)\lesssim C_6$ and $|N_G(P)|=420/15=28$ yield that $|C_G(P)|=14$. Since $Z(C_G(P))⩾P\simeq Z_7$ and $C_G(P)/Z(C_G(P))$ is cyclic, by GZ-Theorem $C_G(P)$ is abelian. It deduces that $C_G(P)$ is cyclic. Assume that $x\in G$ is an element of order $14$. Since $[G:N_G(P)]=15$, there is an injective map $\phi :G\to S_{15}$. Note that $\phi (x)$ is a cycle of length $14$, then ${\phi }^{-1}(\phi (G)\cap A_{15})⊲G$ and so $G$ is not simple. \end{proof}

Proposition

Group of order $264=2^3\cdot 3\cdot 11$ is not simple.

\begin{proof} Assume $G$ is simple, then $n_{11}=12$. Then $|N_G(P)|=22$ and $N_G(P)/C_G(P)\lesssim C_{10}$ yield that $|C_G(P)|=11$. Thus $C_G(P)=P$. Note that $P$ acting on $\Omega :=\{N_G(P)g:g\in G\}$ has two orbits, as $|\Omega |=12$, $P$ is a $11$-subgroup and $P$ has only one fixed point. Take $x\in N_G(P)\setminus C_G(P)$, then $x$ acts on $\{N_G(P)g:g\in G\}$ has at most $2$ fixed points by ^x4xxhg.

Since $o(x)=2$, $\phi (x)\in \mathrm{Sym}(\Omega )$ is a product of $2$-cycles. Notice that $x\in N_G(P)$ fixes $N_G(P)\in \Omega$ and so $x$ is a product of five $2$-cycles. Therefore, $\phi (G)\cap A_{12}\ne \phi (G)$ amd so $G$ has a normal subgroup of index $2$. \end{proof}