By Yakov Berkovich, Zvonimir Janko

ISBN-10: 3110207176

ISBN-13: 9783110207170

This is often the final of 3 volumes of a accomplished and straightforward remedy of finite p-group idea. subject matters lined during this quantity: (a) impression of minimum nonabelian subgroups at the constitution of p-groups, (b) type of teams all of whose nonnormal subgroups have a similar order, (c) levels of irreducible characters of p-groups linked to finite algebras, (d) teams coated via few right subgroups, (e) p-groups of point breadth 2 and subgroup breadth 1, (f) detailed variety of subgroups of given order in a metacyclic p-group, (g) gentle subgroups, (h) p-groups with a maximal trouble-free abelian subgroup of order pÂ², (i) p-groups generated by way of definite minimum nonabelian subgroups, (j) p-groups during which yes nonabelian subgroups are 2-generator. The ebook includes many dozens of unique routines (with tricky routines being solved) and a listing of greater than one thousand learn difficulties and subject matters.

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**Extra info for Groups of Prime Power Order: Volume 3**

**Example text**

Let G be as in (MA1) or (MA2). Write A D hai, B D hbi. (i) Let G be nonmetacyclic as in (MA1). The subgroups A and B are neither G-invariant nor conjugate. The normal closures AG D A G 0 , B G D B G 0 are of order pjAj, pjBj, respectively. Suppose that m > 1. G/ D B E, where E Š Ep2 . Let L < E be of order p such that B L ¤ B G 0 . G/ > 2, a contradiction. Similarly, we consider the case n D 1 so that jGj D p 3 . p 3 / is nonabelian of order p 3 and exponent p > 2). G/ D 1). G/ Š Ep2 , every nonnormal subgroup of G being abelian is cyclic.

It remains to show that such G satisﬁes the hypothesis. In view of H \ A D ¹1º, we obtain that jH j Ä jG W Aj D p n . G/, we have jH j D p , and G is as in (b2). Now let G be nonmetacyclic as in (a2) so m D n. Assume that m > 1. A/ is nonnormal and noncyclic, a contradiction. Thus G is as in (b1). (c) Suppose that all nonnormal cyclic subgroups of G of same order are conjugate; then these subgroups have the same order. Let G be nonmetacyclic as in (MA1) and let m; n > 1. Assume that m > n. A/ D hxi.

G/ belong to p distinct G-classes, and we conclude that p D 2. G/ 2 C 1 > 2, a contradiction. A/ Š Ep2 . Now it follows from (i) and (ii) that C is cyclic. A/ < C . a/ (note that hap i < C and C is cyclic). Write F D A0 C0 . Then F D A0 K, where jKj D p (basic theorem on abelian p-groups). It follows from F 6Ä A that K 6Ä A. Since jKj D p, we get a contradiction to the result of the previous paragraph. 7 (O. Schmidt [Sch4]). G/ D 2, then either p D 2 and G 2 ¹D8 ; Q16 ; H2;m º or G Š Mpn Cq , where q ¤ p are any primes.

### Groups of Prime Power Order: Volume 3 by Yakov Berkovich, Zvonimir Janko

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