## Basic properties of the Thompson subgroup

For any finite group $P$ let $d(P)$ be the minimum number of generators of $P$. (For example, $d(P) = 1$ if and only if $P$ is a nontrivial cyclic group, and $d(Q_8) = 2$.) Let $m(P)$ be the maximum of the integers $d(A)$ as $A$ ranges over all abelian subgroups of $P$. (For example, $m(Q_8) = 1$ and $m(D_8) = 2$.) Define $J(P) = \langle A \ |\ A \leq P\ \mathrm{is\ abelian\ and}\ d(A) = m(P) \rangle$. ($J(P)$ is called the Thompson subgroup of $P$.)

1. Prove that $J(P)$ is characteristic in $P$.
2. For each of the groups $P$ in the list $Q_8$, $D_8$, $D_{16}$, and $QD_(16)$, give the abelian subgroups $A$ which satisfy $d(A) = m(P)$. [Hint: Use the lattices of subgroups which were given in §2.7.]
3. Show that $J(Q_8) = Q_8$, $J(D_8) = D_8$, $J(D_16) = D_16$, and $J(QD_{16}) \cong D_8$.
4. Prove that if $Q \leq P$ and $J(P) \leq Q$, then $J(Q) = J(P)$. Deduce that if $P$ is a subgroup of a finite group $G$ and if $J(P) \leq Q$ for some $Q \vartriangleleft G$, then $J(P) \vartriangleleft G$.

1. First we prove some lemmas.

Lemma 1: Let $P$ be a group, $A \leq P$ a subgroup, and $\varphi \in \mathsf{Aut}(P)$ an automorphism. Then $d(A) = d(\varphi[A])$. Proof: Suppose $S$ is a minimal generating set for $A$; that is, $A = \langle S \rangle$ and $|S| = d(A)$. Note that $\varphi[A] = \varphi[\langle S \rangle] = \langle \varphi[S] \rangle$. Since $\varphi$ is bijective, $|\varphi[S]| = |S|$. Thus $d(\varphi[A]) \leq d(A)$. Now $\varphi^{-1}$ is also an automorphism of $G$, so that $d(A) = d(\varphi^{-1}[\varphi[A]]) \leq d(\varphi[A])$. Hence $d(A) = d(\varphi[A])$. $\square$

Now to the main result. Suppose $\varphi \in \mathsf{Aut}(P)$. Then $\varphi[J(P)] = \varphi[\langle \{ A \ |\ A \leq P\ \mathrm{is\ abelian\ and}\ d(A) = m(P) \} \rangle]$ $= \langle \{ \varphi[A] \ |\ A \leq P\ \mathrm{is\ abelian\ and}\ d(A) = m(P) \} \rangle$. Now $d(\varphi[A]) = d(A)$. Moreover, as $A$ ranges over all abelian subgroups of $P$, so does $\varphi[A]$. Thus $\varphi[J(P)] = J(P)$. Hence $J(P)$ is characteristic in $P$.

1. Every proper subgroup of $Q_8$ is abelian. Moreover, every proper subgroup of $Q_8$ is cyclic; thus $m(Q_8) = 1$, and every proper subgroup $H$ satisfies $d(H) = m(Q_8)$. Specifically, these subgroups are $\langle -1 \rangle$, $\langle i \rangle$, $\langle j \rangle$, and $\langle k \rangle$.
2. Every proper subgroup of $D_8$ has order 2 or 4 and thus is abelian. Six of these subgroups are cyclic, and the remaining two are noncyclic groups of order 4, hence isomorphic to $V_4$. Thus $m(D_8) = 2$, and the subgroups $H$ such that $d(H) = m(D_8)$ are $\langle s,r^2 \rangle$ and $\langle rs,r^2 \rangle$.
3. Every proper subgroup of $D_{16}$ of order 2 or 4 is abelian. Ten of these are cyclic, and the remaining four are noncyclic groups of order 4, so that $m(D_{16}) \geq 2$. Of the three subgroups of order 8, two are nonabelian and one is cyclic. Hence $m(D_{16}) = 2$. The subgroups $H$ which satisfy $d(H) = m(D_{16})$ are $\langle sr^2, r^4 \rangle$, $\langle s,r^4 \rangle$, $\langle sr^3,r^4 \rangle$, and $\langle sr^5,r^4 \rangle$.
4. Every proper subgroup of $QD_{16}$ of order 2 or 4 is abelian. Eight of these are cyclic, and the remaining two are noncyclic groups of order 4, so that $m(QD_{16}) \geq 2$. Of the three subgroups of order 8, two are nonabelian and one is cyclic. Hence $m(QD_{16}) = 2$. The subgroups $H$ which satisfy $d(H) = m(QD_{16})$ are $\langle \sigma^4, \tau \rangle$ and $\langle \sigma^2\tau, \sigma^4 \rangle$.
2. From the subgroup lattices of these groups, we clearly have $J(Q_8) = Q_8$, $J(D_8) = D_8$, $J(D_{16}) = D_{16}$, and $J(QD_{16}) = \langle \sigma^2, \tau \rangle$. Note that in $QD_{16}$, $(\sigma^2)^4 = \tau^2 = 1$ and $\sigma^2 \tau = \tau (\sigma^2)^{-1}$. Thus $\langle \sigma^2, \tau \rangle \cong D_8$.
3. We begin with a lemma.

Lemma 2: If $Q \leq P$ then $m(Q) \leq m(P)$. Proof: Let $\mathcal{A}(G)$ denote the set of abelian subgroups of $G$. Clearly we have $\mathcal{A}(Q) \subseteq \mathcal{A}(P)$. Thus $m(Q) \leq m(P)$. $\square$

Lemma 3: If $J(P) \leq Q \leq P$ then $m(Q) = m(P)$. Proof: By Lemma 2 we have $m(Q) \leq m(P)$. Moreover, there exists $A \leq Q$ such that $d(A) = m(P)$; hence $m(Q) = m(P)$. $\square$

Now to the main result. If $J(P) \leq Q \leq P$, then $m(Q) = m(P)$. If $A \leq P$ is abelian and $d(A) = m(P)$, then $A \leq Q$ (Since $A \leq J(P)$) and $d(A) = m(Q)$. Thus $J(P) \leq J(Q)$. Likewise, if $A \leq Q$ is abelian and $d(A) = m(Q)$, then $A \leq P$ and $d(A) = m(P)$. Hence $J(Q) = J(P)$.

Suppose now that $G$ is a finite group with $P \leq G$. Suppose further that $J(P) \leq Q$ for some normal subgroup $Q \leq G$. Now $J(P) = J(Q)$ is characteristic in $Q$ by part 1, so that $J(P)$ is normal in $G$.