# A question on the basic properties of Thompson subgroup of a group

For any finite group $P$ let $d(P)$ be the minimum number of generators of $P$. Let $m(P)$ be the maximum of the integers $d(A)$ as $A$ runs over all abelian subgroups of $P$. Define
$J(P)=\langle A$ s.t. $A$ is an abelian subgroup of $P$ with $d(A)=m(P)\rangle$, called the Thompson subgroup of $P$. (Sec. 4.4, Ex. 20 of Abstract Algebra: Dummit & Foote)

For example, let $\operatorname{QDih}=\langle r,s$ s.t. $r^8=s^2=1, rs=sr^3\rangle$ be the quasidihedral group of order $16$. Then $d(\operatorname{QDih})=2$. All but three subgroups of $\operatorname{QDih}$ are abelian. They are $\langle s,r^2\rangle\cong \operatorname{Dih}(4)$ dihedral group, $\langle sr,r^2\rangle \cong \operatorname{Q}(8)$ quaternion group, and $\operatorname{QDih}$ itself. $d(\langle s,r^4\rangle)=d(\langle sr,r^4\rangle)=2$, the maximum possible, and so $m(\operatorname{QDih})=2$. $$J(\operatorname{QDih})=\langle\langle s,r^4\rangle,\langle sr,r^4\rangle\rangle=\langle s,r^2\rangle\cong \operatorname{Dih}(4)$$ If you're still confused, you may check that $J(\operatorname{Alt}(4))=\langle(12)(34),(13)(24)\rangle$, the unique subgroup of order $4$.

I want to :

-prove that $J(P)\unlhd^{\operatorname{char}} P$,
-and that if $J(P)\le Q\le P$, then $J(P)=J(Q)$. Deduce that if $P\le G$ and $J(P)\le Q\le P$ s.t. $Q\unlhd G$, then $J(P)\unlhd G$.

There are only a few useful examples to illustrate these facts and give us useful information. In reality, $J(\operatorname{Q}(8))=\operatorname{Q}(8)$, $J(\operatorname{Dih}(4))=\operatorname{Dih}(4)$, and $J(\operatorname{Dih}(8))=\operatorname{Dih}(8)$. I call them 'useless' groups.

Regarding the first statement, what I think is that since for all $\sigma\in \mathrm A\mathrm u\mathrm t(P)$, $\sigma$ preserves the subgroup structure, and the min. no. of generators a subgroup has, if $A$ is an abelian subgroup of $P$ with $d(A)=m(P)$, then $d(\sigma(A))=m(P)$ also. So in some sense, $\sigma$ 'permutes' the abelian subgroups of $P$ which have max. of min. no. of generators possible. So $\sigma(J(P))=<\sigma(A)>=J(P)$. i.e. $J(P)\unlhd^{\operatorname{char}} P$(?)

Suppose it's true that if $J(P)\le Q\le P$, then $J(P)=J(Q)$.
If $P\le G$ and $J(P)\le Q\le P$ s.t. $Q\unlhd G$, then $J(Q)\unlhd^{\operatorname{char}}Q\unlhd G$. So $J(P)=J(Q)\unlhd G$.

But I couldn't prove the 'if $J(P)\le Q\le P$, then $J(P)=J(Q)$' in the first place since there's no 'useful' group for me to work with. Can somebody help?

• Where you wrote things like $<s, r^2>$ instead of $\langle s,r^2\rangle,$ couldn't you have just googled something like "latex symbols" to find out how to code that? – Michael Hardy Jun 22 '17 at 7:07
• LOL... I ready thought that the $\langle \dots \rangle$ and $<\dots>$ were the same thing. – user441558 Jun 22 '17 at 7:26
• It follows easily from the definition. If $J(P) \le Q \le P$, then all of the abelian subgroups $A$ of $P$ with $d(A)=m(P)$ are contained in $J(P)$ and hence in $Q$, so $J(Q)=J(P)$. The final part is also easy. If $Q \unlhd G$ then $J(Q)=J(P) \unlhd G$, because $J(Q) {\rm char} Q$. – Derek Holt Jun 22 '17 at 7:53

Observe that $J(P) \subseteq Q$ is given. It means that it is generated by Abelian groups of $Q$ which has $d(A)=m(P)$. But what is $m(Q)$? $m(Q)=max \{\ d(A) |A$runs over all abelian subgroups of Q $\}\$. Also $m(P) \geq m(Q)$. This implies that those abelian subgroups that generate $J(P)$ for those we have $d(A)=m(Q)=m(P)$. Hence $J(P)$ and $J(Q)$ are generated by same Abelian subgroups of Q and hence $J(P)=J(Q)$.
• LOL... My attempt: Suppose $J(P)\le Q\le P$. $J(P)$ is essentially the subgroup of $P$ generated by the abelian subgroups of $P$, the abelian subgroups of which can only be generated by the max. no. of generators possible among all the abelian subgroups of $P$. Then those abelian subgroups are all contained in $Q$. So $m(Q)=m(P)$ and $J(Q)$ is generated by those abelian subgroups of $Q$ which can only be generated by the max. no. of generators possible among all the abelian subgroups of $Q$. The abelian subgroups of $Q$ are exactly those which generate $J(P)$. – user441558 Jun 22 '17 at 8:09