More on a question of M. Newman on isomorphic subgroups of solvable groups

• Published in: Journal of algebra 2019-08, Vol.532, p.1-7
• Main Authors: Glauberman, George, Robinson, Geoffrey R.
• Format: Article
• Language: English
• Subjects: Maximal subgroups; Solvable groups
• ISSN: 0021-8693; 1090-266X
• DOI: 10.1016/j.jalgebra.2019.05.007

We make further remarks on a question of Moshe Newman, which asked whether it is the case that if H and K are isomorphic subgroups of a finite solvable group G and H is maximal in G, then K is also maximal. This continues work begun in [2] by I.M. Isaacs and the second author. We prove here that if Newman's question has a negative answer for the triple (G,H,K) (i.e. H is maximal in G, but the isomorphic subgroup K is not), and [G:H] is a power of the prime p, then p≤3 and, for q=5−p, we haveOq′(H)=Oq′(K)=Oq′(G), and Newman's question also has a negative answer for the triple (G⁎,H⁎,K⁎), where G⁎=G/Oq′(G), etc. Furthermore, we prove that G has a homomorphic image G¯ such that Newman's question has a negative answer for the triple (G¯,H¯,K¯), while F(G¯),F(H¯) and F(K¯) are all q-groups, and O{2,3}(H¯) involves Qd(q). As an application, we prove that if G is a finite solvable group such that H and K are isomorphic subgroups of G with H maximal and K not maximal, with [G:H]=[G:K] a power of the prime p, then p≤3 and a Hall {2,3}-subgroup L of H necessarily involves S3,A4 and a non-Abelian group of order 8 (in fact, L involves at least one of S4 or Qd(3)). In particular, L is neither 2-closed nor 3-closed.