Exercise 4.4.20 (Thompson subgroup of $P$)

Question

For any finite group $P$ let $d(P)$ be the minimum number of generators of $P$ (so, for example, $d(P) = 1$ if and only if $P$ is a non trivial cyclic group and $d(Q_8) = 2$). Let $m(P)$ be the maximum of the integers $d(A)$ as $A$ runs over the {\bf abelian} subgroups of $P$ (so, for example, $m(Q_8) = 1$ and $m(D_8) = 2$). Define
$$J(P) = \langle A \mid A 	ext{ is an abelian subgroup of $P$ with $d(A) = m(P)$} angle.$$
($J(P)$ is called the {\bf Thompson subgroup} of $P$.)
\begin{enumerate}
\item[{\bf(a)}] Prove that $J(P)$ is a characteristic subgroup of $P$.
\item[{\bf(b)}]For each of the following groups $P$ list all abelian subgroups $A$ of $P$ that satisfy $d(A) = m(P)$: $Q_8,\ D_8,\ D_{16}$ and $QD_{16}$ (where $QD_{16}$ is the quasidihedral group of order $16$ defined in Exercise 11 of Section 2.5). [Use the lattices of subgroups for these groups in Section 2.5.]
\item[{\bf(c)}]Show that $J(Q_8) = Q_8,\ J(D_8) =D_8,\ J(D_{16} ) = D_{16}$ and $J(QD_{16})$ is a dihedral subgroup of order $8$ in $QD_{16}$.
\item[{\bf(d)}]Prove that if $Q\leq P$ and $J(P)$ is a subgroup of $Q$, then $J(P) = J(Q)$. Deduce that if $P$ is a subgroup (not necessarily normal) of the finite group $G$ and $J(P)$ is contained in some subgroup $Q$ of $P$ such that $Q \unlhd G$, then $J(P) \unlhd G$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\begin{proof} Let $J(P)$ be the Thompson subgroup of $P$.
\begin{enumerate}
\item[{\bf(a)}] Consider an automorphism $\sigma \in \mathrm{Aut}(P)$. We must prove $\sigma(J(P)) = J(P)$.

Note first that for every abelian subgroup of $P$, $\sigma(A)$ is an abelian subgroup of $P$ (since $\sigma$ is an automorphism), and $d(\sigma(A)) = d(A)$: if $A = \langle a_1, a_2,\ldots,a_mangle$, where $m = d(A)$, then $\sigma(A) = \langle \sigma(a_1), \sigma(a_2), \ldots, \sigma(a_m)angle$, thus $d(\sigma(A)) \leq m$. If $\sigma(A) = \langle b_1,b_2, \ldots,b_l angle$, where $b_i \in P$ for all $i$, then 
$$A = \sigma^{-1}(\sigma(A)) = \langle \sigma^{-1}(b_1),\ldots, \sigma^{-1}(b_l) angle.$$ Hence $m = d(A) \leq l$, so $m$ is the minimum number of generators of $\sigma(A)$, i.e., $m = d(\sigma(A))$. This proves that for all abelian subgroups $A$,
\begin{align}
d(\sigma(A)) = d(A).
\end{align}

\bigskip
By definition, $J(P) = \langle A_1,A_2,\ldots,A_kangle$, where $E =  \{A_1,A_2,\ldots,A_k\}$ is the set of abelian subgroups of $P$ such that $d(A_i) = m(P)$. Then 
$$\sigma(J(P)) = \langle \sigma(A_1),\sigma(A_2), \ldots, \sigma(A_k)angle,$$
 where $\sigma(A_i)$ is an abelian subgroup for all $i$,  Moreover, by (1),  $d(\sigma(A_i)) = d(A_i) = m(P)$. Since $\sigma(A_i) 
e \sigma(A_j)$ if $i
e j$, $\sigma$ permutes the elements of $E$, so
 $$\{A_1,A_2,\ldots,A_k\} = \{\sigma(A_1),\sigma(A_2),\ldots \sigma(A_k)\}.$$
 Therefore 
 $$\sigma(J(P)) = \langle \sigma(A_1),\sigma(A_2), \ldots, \sigma(A_k)angle =  \langle A_1,A_2,\ldots,A_kangle =J(P),$$
 so
 $$\sigma(J(P)) = J(P).$$
 Since this is true for every automorphism $\sigma \in \mathrm{Aut}(P)$, this shows that $J(P)$ is a characteristic subgroup of $P$.

\item[{\bf(b)}]
\begin{itemize}
\item If $P = Q_8$, then the lattice of subgroups of $P$ is 

  \bigskip    
      \begin{center}
\begin{tikzpicture}

ode (1b)   at (0,0) {$\langle 1 angle$};

ode (-1b)   at (0,2) {$\langle-1angle$};

ode (ib)   at (-2,4) {$\langle i angle$};

ode (jb)   at (0,4) {$\langle j angle$};

ode (kb)   at (2,4) {$\langle k angle$};

ode (Gb)   at (0,6) {$Q_8$};
\draw (1b) edge  (-1b)  ;
\draw (-1b) edge  (ib)  ;
\draw (-1b) edge  (jb)  ;
\draw (-1b) edge  (kb)  ;
\draw (Gb) edge  (ib)  ;
\draw (Gb) edge ( jb)  ;
\draw (Gb) edge  (kb)  ;
\end{tikzpicture}
\end{center}

The abelian subgroups of $Q_8$ are $\langle -1 angle, \langle i angle,\langle j angle,\langle k angle$, all cyclic, so $m(Q_8) = 1$, and the set $S_{Q_8}$ of abelian subgroups $A$ of $Q_8$ such that $d(A) = m(Q_8) = 1$ is
\begin{align}
S_{Q_8} = \{\langle -1 angle, \langle i angle,\langle j angle,\langle k angle\}.
\end{align}



\item If $P = D_8$, then the lattice of subgroups of $P$ is 
		\begin{center}
\begin{tikzpicture}
    
ode (n0) at (3,0) {$\langle 1 angle$};
    
ode (n1) at (0,1.5) {$\langle s r angle$};
    
ode (n2) at (1.5,1.5) {$\langle  sr^3 angle$};
    
ode (n3) at (3,1.5) {$\langle r^2angle$};
    
ode (n4) at (4.5,1.5) {$\langle  sr^2angle$};
    
ode (n5) at (6,1.5) {$\langle  sangle$};
    
ode (n6) at (1.5,3) {$\langle r^2, s r angle$};
    
ode (n7) at (3,3) {$\langle r angle$};
    
ode (n8) at (4.5,3) {$\langle r^2,   s  angle$};
    
ode (n9) at (3,4.5) {$\qquad \ \   \langle r,sangle = D_8$};
    \draw (n0) edge (n3) edge (n2) edge (n1) edge (n4) edge (n5);
    \draw (n6) edge (n1) edge (n2) edge (n3);
    \draw (n8) edge (n3) edge (n4) edge (n5);
    \draw (n3) edge (n7);
    \draw (n9) edge (n6) edge (n7) edge (n8);

\end{tikzpicture}
	\end{center}
	
	Then $m(D_8) = 2$, and the set $S$ of abelian subgroups $A$ of $D_8$ such that $d(A) = m(D_8) = 2$ is
	\begin{align}
	S_{D_8} = \{\langle r^2, sr angle, \langle r^2,s angle\},
	\end{align}
	where $\langle r^2, sr angle \simeq \langle r^2,s angle\} \simeq V_4$.
	
\item If $P = D_{16}$, then the lattice of subgroups of $P$ is given p. 70, which shows that $m(D_{16}) = 2$. All subgroups of order $4$ are abelian.

Since $s r^2 = r^6 s 
e r^2 s$, the subgroup $\langle s, r^2 angle$ is not abelian. Similarly, since $(sr) r^2 = sr^3 = r^5 s 
e rs = r^2(sr)$, the subgroup $\langle sr, r^2 angle$ is not abelian.

The set $S_{D_{16}}$ of abelian subgroups $A$ of $D_{16}$ such that $d(A) = m(D_{16}) = 2$ is
\begin{align}
S_{D_{16}} = \{\langle sr^2, r^4 angle, \langle s,r^4 angle, \langle sr^3, r^4 angle, \langle sr^5, r^4 angle\}.
\end{align}
(These four subgroups are isomorphic to $V_4$, so are not cyclic.)


\item If $P = QD_{16}$, then the lattice of subgroups of $P$ is


      \bigskip    
      \begin{center}
\begin{tikzpicture}

ode (1)   at (0,0) {$\langle 1 angle$};

ode (2)   at (0,2) {$\langle \sigma^4angle$};

ode (3)   at (-1.7,2) {$\langle	au angle$};

ode (4)   at (-3.4,2) {$\langle  	au\sigma^4 angle$};

ode (5)   at (-5.1,2) {$\langle 	au \sigma^6 angle$};

ode (6)   at (-6.8,2) {$\langle 	au \sigma^2 angle$};

ode (7)    at (5,4) {$\langle 	au \sigma^7 angle$};

ode (8)    at (3,4) {$\langle 	au \sigma angle$};

ode (9)    at (0,4) {$\langle \sigma^2 angle$};

ode (10)    at (-3,4) {$\langle \sigma^4, 	au  angle$};

ode (11)    at (-6,4) {$\langle \sigma^4, 	au  \sigma^2 angle$};

ode (12)    at (3,6) {$\langle \sigma^2, 	au \sigma angle$};

ode (13)    at (0,6) {$\langle \sigma angle$};

ode (14)    at (-3,6) {$\langle \sigma^2, 	au  angle$};

ode (15)    at (0,8) {$QD_{16}$};

\draw (1) edge  (2) edge (3) edge (4) edge (5) edge (6);
\draw (2) edge  (7) edge (8) edge (9) edge (10) edge (11);
\draw (11) edge (5) edge (6);
\draw (10) edge (3) edge (4);
\draw (9) edge  (13);
\draw (15) edge  (13) edge (12) edge (14);
\draw (14) edge  (9) edge (10) edge (11);
\draw (12) edge  (9) edge (8) edge (7);
\end{tikzpicture}
\end{center}
\end{itemize}
Then $m(QD_{16}) = 2$.

All subgroups of order $4$ are abelian.

By Exercise 2.5.11, 
$$\sigma^8 = 	au^2 = 1,\qquad \sigma 	au = 	au \sigma^3,$$
Therefore
$$\sigma^2 	au = \sigma (\sigma 	au) = \sigma 	au \sigma^3 = 	au \sigma^3 \sigma^3 = 	au \sigma^6 
e 	au \sigma^2,$$
thus $\langle \sigma^2, 	auangle$ is not abelian.

Similarly,
$$\sigma^2 (	au \sigma) = 	au \sigma^7 
e 	au \sigma^3 = (	au \sigma) \sigma^2,$$
thus $\langle \sigma^2, 	au \sigma angle$ is not abelian.
The set $S_{QD_{16}}$ of abelian subgroups $A$ of $QD_{16}$ such that $d(A) = m(QD_{16}) = 2$ is
\begin{align}
S_{QD_{16}} = \{\langle \sigma^4, 	au \sigma^2 angle, \langle \sigma^4, 	au angle\}.
\end{align}

\item[{\bf(c)}]
\begin{itemize}
\item $P = Q_8$.

By (2),
\begin{align*}
J(Q_8) &= \langle  \langle -1 angle, \langle i angle,\langle j angle,\langle k angle\
&= \langle -1, i, j ,k  angle\
&= Q_8,
\end{align*}
since $Q_8 = \langle i, j angle$. So
$$J(Q_8) = Q_8.$$

\item $P = D_8$.

By (3),
\begin{align*}
J(D_8) &= \langle \langle r^2, sr angle, \langle r^2,s angle angle\
&= D_8\
\end{align*}
Since the smallest subgroup containing $\langle  r^2, sr angle $ and $\langle r^2,s angle$ is $D_8$ (see the lattice of subgroups of $D_8$ above). So
$$J(D_8) = D_8.$$

\item $P = D_{16}$.

By (4), and the lattice of subgroups of $D_{16}$ p. 70,
\begin{align*}
J(D_{16}) &= \langle \langle sr^2, r^4 angle, \langle s,r^4 angle, \langle sr^3, r^4 angle, \langle sr^5, r^4 angleangle\
&=\langle  \langle s,r^2 angle, \langle sr, r^2 angle angle\
&=D_{16},
\end{align*}
So
$$J(D_{16}) = D_{16}.$$

\item $P = QD_{16}$.

By (5), and the lattice of subgroups of $QD_{16}$,
\begin{align*}
J(QD_{16}) &= \langle \langle \sigma^4, 	au \sigma^2 angle, \langle \sigma^4, 	au angleangle\
&= \langle \sigma^2, 	au angle.
\end{align*}
where $\langle \sigma^2, 	au angle$ is a subgroup of $QD_{16}$ of order $8$. We have proved in Exercise 2.5.11, part 2° (b), that
$$\langle \sigma^2, 	au angle = \{1,\sigma^2, \sigma^4, \sigma^6, 	au , 	au \sigma^2, 	au \sigma^4, 	au \sigma^6\},$$
where $(\sigma^2)^4 =	au^2 = 1$ and $	au \sigma^2 = (\sigma^2)^3 	au$ is isomorphic to $D_8$. So
$$J(QD_{16}) \simeq D_8.$$
\end{itemize}

\item[{\bf(d)}]  Since the set of subgroups of $P$ is finite, there is some abelian subgroup $A_0$ of $P$ such that $d(A_0) = m(P)$. Then $A_0 \leq J(P)$ by definition of $J(P)$. Therefore $m(J(P))$, which is the maximum of $d(A)$ when $A$ runs over all abelian subgroup of $J(P)$ is at least equal to $m(P)$: $m(J(P)) \geq m(P)$. 

Moreover $J(P) \leq P$, thus $m(J(P)) \leq m(P)$. This shows that
$$m(J(P)) = m(P).$$

Suppose now that $Q$ is a subgroup of $G$ such that
$$J(P) \leq Q \leq P.$$
Then $m(J(P)) \leq m(Q) \leq m(P)$, where $m(J(P)) = m(P)$, thus $m(Q) = m(P)$.
If $A$ is an abelian subgroup of $Q$ such that $d(A) = m(Q)$, then $A$ is an abelian subgroup of $P$ such that $d(A) = m(P)$, therefore $A \leq  J(P)$. This shows that $J(P)$ contains the subgroup generated by these subgroups, so $J(Q) \leq J(P)$.

Conversely, if $A$ is an abelian subgroup of $P$ such that $d(A) = m(P)$, then $A \leq J(P)$, thus $A \leq Q$, so $A$ is an abelian subgroup of $Q$ such that $d(A) = m(Q)$, therefore $A \leq J(Q)$. This shows that $J(P) \leq J(Q)$, and so
$$J(P) = J(Q).$$

Suppose now that $J(P) \leq Q  \leq P \leq G$, where $Q \unlhd G$. By the argument above, $J(P) = J(Q)$, thus $J(P)$ is a characteristic subgroup of $Q$. Since $Q \unlhd G$, by Exercise 8 (a), we obtain
$$J(P) \unlhd G.$$
\end{enumerate}
\end{proof}

Exercise 4.4.20 (Thompson subgroup of $P$)

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Exercise 4.4.20 (Thompson subgroup of $P$)

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