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.

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