Auditor group

In mathematics , especially in group theory , for a prime p, each is called a multiplicative group

${\displaystyle \mathbb {C} _{p^{\infty }}=\{\exp(2\pi \mathrm {i} n/p^{m})\mid n\in \mathbb {Z} ,m\in \mathbb {N} \}}$

isomorphic group a p - examiner group or a p - quasi-cyclic group . consists of the complex roots of unity , the order of which is a power of p . ${\displaystyle \mathbb {C} _{p^{\infty }}}$

It is an Abelian , countable group.

According to the definition, the p -tester groups are isomorphic to one another, therefore one simply speaks of the p -tester group without further specification . It is said that a group G is a group of auditors if there is a prime p there, so G a p -Prüfergruppe is. The examiner groups for different prime numbers are not isomorphic.

The examiner groups are named in honor of the mathematician Heinz Prüfer .

Equivalent Definitions

Let p be a prime number and G a group. Each of the following five properties is equivalent to G being a p -assignment group, and each of these properties can therefore be used as the definition of the auditing group.

a) G is isomorphic to the factor group , wherein the of the rational numbers with formed sub-group of designated. ${\displaystyle \mathbb {Z} [1/p]/\mathbb {Z} }$ ${\displaystyle \mathbb {Z} [1/p]}$ ${\displaystyle n/p^{m}}$ ${\displaystyle n\in \mathbb {Z} ,m\in \mathbb {N} }$ ${\displaystyle (\mathbb {Q} ,+)}$

Proof: The homomorphism is surjective and has the core . ${\displaystyle \mathbb {Z} [1/p]\rightarrow \mathbb {C} _{p^{\infty }}:q\mapsto \exp(2\pi iq)}$ ${\displaystyle \mathbb {Z} }$

b) G is isomorphic to the factor group , where F is the free Abelian group (i.e. the free - module ) with a countable infinite base and R is the subgroup of F generated by . ${\displaystyle F/R}$ ${\displaystyle \mathbb {Z} }$ ${\displaystyle \{\ a_{0},a_{1},\ldots ,a_{n},\ldots \}}$ ${\displaystyle \{pa_{0},a_{0}-pa_{1},a_{1}-pa_{2},\ldots ,a_{n}-pa_{n+1},\ldots \}}$

c) G has a presentation

${\displaystyle \langle x_{1},x_{2},\dots |x_{1}^{p}=1,x_{2}^{p}=x_{1},x_{3}^{p}=x_{2},\dots \rangle .}$

Proof: Let L be a free (non-Abelian) group over a countable basis and S the normal divisor generated by . For every natural number j is the canonical image of in . It is clear that of any two of the elements one is a power of the other, that is, they interchange with one another. Since they generate is Abelian, in other words, S contains the commutator group K (L) . According to the second isomorphism, is isomorphic to . Now there is a free, Abelian group (free as an Abelian group) with the images of the elements as a base in and is generated by. Now you close with b). ${\displaystyle \{\ c_{0},c_{1},\ldots ,c_{n},\ldots \}}$ ${\displaystyle \{c_{0}^{p},c_{0}c_{1}^{-p},c_{1}c_{2}^{-p},\ldots ,c_{n}c_{n+1}^{-p},\ldots \}}$ ${\displaystyle x_{j}}$ ${\displaystyle c_{j}}$ ${\displaystyle L/S}$ ${\displaystyle x_{j}}$ ${\displaystyle x_{j}}$ ${\displaystyle L/S}$ ${\displaystyle L/S}$ ${\displaystyle L/S}$ ${\displaystyle (L/K(L))/(S/K(L))}$ ${\displaystyle L/K(L)}$ ${\displaystyle \{\ d_{0},d_{1},\ldots ,d_{n},\ldots \}}$ ${\displaystyle \{\ c_{0},c_{1},\ldots ,c_{n},\ldots \}}$ ${\displaystyle L/K(L)}$ ${\displaystyle S/K(L)}$ ${\displaystyle \{d_{0}^{p},d_{0}d_{1}^{-p},d_{1}d_{2}^{-p},\ldots ,d_{n}d_{n+1}^{-p},\ldots \}}$

d) G has a generating system such that , and for all . ${\displaystyle \ (a_{n})_{n\in \mathbb {Z} }}$ ${\displaystyle \ a_{0}\not =1}$ ${\displaystyle \ a_{0}^{p}=1}$ ${\displaystyle \ a_{n+1}^{p}=a_{n}}$ ${\displaystyle n\geq 0}$

e) G is the association of an ascending order , where C _n for each index n a cyclic group of order p ⁿ is. ${\displaystyle C_{0}\leq C_{1}\leq \ldots \leq C_{n}\leq \ldots }$

properties

• Every real subgroup of a reviewer group is cyclic and especially finite . The panel has for each number n just a subgroup of order p ⁿ . The number of subgroups of an examiner group is well-ordered through the inclusion . The examiner group is therefore not essential as a module . ${\displaystyle \mathbb {Z} }$

• An infinite Abelian group is a reviewer group if and only if it is isomorphic to every factor group according to a real subgroup.

• The examiner groups are divisible . Their meaning is evident from the following sentence:

Every divisible, Abelian group is isomorphic to a (finite or infinite) direct sum , in which every summand is a tester group or isomorphic to the additive group of rational numbers.

For example, the additive group is the direct sum of its p - Sylow groups , which are nothing other than the p - tester groups. ${\displaystyle \mathbb {Q} /\mathbb {Z} }$

Individual evidence

1. ↑ J. Calai: Éléments de théorie des groupes , Chapter IV, Exercise. 34, page 172
2. ^ DJS Robinson : A Course in the Theory of Groups , Springer-Verlag 1996, ISBN 0-387-94461-3 , page 94: Quasicyclic Groups
3. ^ JJ Rotman: An Introduction to the Theory of Groups , 4th edition 1999, sentence 10.13 and exercise 10.5
4. ↑ J. Calais: Éléments de théorie des groupes , Presses universitaires de France, Paris 1984, Chapter IV, Exercise 34, page 172
5. ^ B. Baumslag et B. Chandler, Group Theory , Mc-Graw Hill, 1968, Theorem 6.31, p. 206
6. ^ That every group of examiners has this quality can be found in J. Calais: Éléments de théorie des groupes , Presses universitaires de France, Paris 1984, Chapter IV, Exercise. 34, f), page 172. For the reversal, see JJ Rotman: An Introduction to the Group Theory , 4th edition 1999, exerc. 10.40, iii, p. 330
7. ^ JJ Rotman: An Introduction to the Group Theory , 4th edition 1999, sentence 10.28, page 323
8. ^ DJS Robinson: A Course in the Theory of Groups , Springer-Verlag 1996, ISBN 0-387-94461-3 , sentence 4.1.5