Белоногов. Задачник по теории групп

G

#$ G

A ! RG(G) G : ! ( .6

. " g Lg := (X → X | x → g−1x) SX

LG := (G → SX | g → Lg)

G

G 0 RG LG :

% CSX (RG(G)) = LG(G) CSX (LG(G)) = RG(G)

" ! > H ≤ G X E

G H

( " g G RH,g := (X → X | Ht → Htg (t G)) SX

RG,H := (G → SX | g → RH,g)

G

G &

H

L RG,H (G) Ker(RG,H )

. K ≤ G >

RG,H : RG,K H K G.

0 G :

" RG,K # K G F !

G ≤ SX ! G RG,K (G)! K = Gx x X

" " ( H := a, b ! a = (12345) b = (2354) > H E .

./ S5! K := a, b2 E

(/ A5

. + ( # S6 " ! , " S5!A6 E " ! , " A5

" # H ≤ G X = {H}G E ! 4

H

( " g G Cg := (X → X | Ht → Htg (t G)) SX

CG,H := (G → SX | g → Cg)

G

. CG,H : " RG,NG(H)

"$ H E n G ( G/HG ≤ Sn A ! |G/HG| n! |H : HG|

(n − 1)!

. I G ! |G| n!

2

" n = 4

Sn ! n!

AnD

An ! n

" n = 1 x {1, . . . , n}

( (Sn)x E Sn

. (An)x E An

" ! 4 "

! " .

" I G H

p HG = 1!

p G!

p2 |G|!

CG(P ) = P " # # p P G

" H E p G HG = 1

" I G & 5

: 5! G

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!

( |G : H| = 7, o (a) > 12D

. |G : H| = 8, o (a) > 15D 0 |G : H| = 9, o (a) > 20D % |G : H| = 10, o (a) > 30D

& |G : H| = n, o (a) = p > n! p E n N

" " ( I G hf H h

f ≤ p! p E # # h! H G

. ) ( # # G

" # G E ! P E 4

g E : n G > P(g)

$! # n

"$ G E 4 4 . P $ > G " N "!

G = N P

" G = AB, A B E G, A ∩ B = 1 |A| = 2n 4 n > G "

2

" G = AB! A B E G! |A ∩ B|

! . T A $

> G " N "! G = N T

" I |G| = pnm! m ≤ 2p p E m, n N!

G " pn pn−1

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F E , 4 " W ( J # W $# #

# E(n)

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% E(1) D(R+)

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z = x + iy > O ϕ #

, # z → eiϕz!

−−→

Ox E , # z → z! OM ! M = (a, b)! E , # z → z + c! c = a + ib

( : 4 #

, H

z → eiϕz + c z → eiϕz + c,

0 ≤ ϕ < 2π c C (z C)

. 2 # : E(2)

"$ F E , $

O + ! # , F

O # !

O

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E(2) ! , F

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> H G E : {A, B}

G ! G = AB (= BA) I G = AB!

A ≤ G, B ≤ G A ∩ B = 1! ! B

G A A G

B G! " G

G

7 G 2!

G H "! H ∩ Hx = 1 x G \ H

! NG(H) = H)D

G " N "! G = N H

N 4 !

H E 4 % G

9 '.. ! # G

A 8 %8

4 # #

G H G ! H E G

# !

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,$

# L ,$ Q8 D8.

# ) ,$ Zn " n N. L Zn

# G = AB! A B E G

( I A E ! 4 A! B E !

4 B! G = A B D 4 ! A ∩ B = 1! A ∩ B = 1D

. I A1 A G

B1 B!

A ∩ B

# < # D6, D10, D30, A4 ,

$ " 4 #

#

# 2

4

# ! H E m # G. I m

H, H G.

# " G = AB! G E A, B E 4

! |A| ≤ pq! p E #

# |A|! q E # # |B| >

NG(B) = B! B G

# # A, B, C E G

( A B C > A B, C F A B ∩ CO

. A(B C) = AB AC

0 (B C)A = BA CA

% ; . 0 !

∩.

#$ A B E G ; H ( AB ≤ G!

. AB = BA!

0 " a A b B Ab ∩ Ba =

# I G = AB! A B E G!

|A : A ∩ B| = |G : B|.

# A, B, C E G ! G = AB = AC F G = A(B ∩ C)O

# 2 # G! " " A B !

( AB E G! BA # G!

. |AB| = |BA|!

# H1 , H2 , H3 E ! H1H2H3 = H2HiHj = H3HsHt i, j, s, t {1, 2, 3} > H1H2H3 E

G

# A, B, C E G! 4

>

AB = BC = CA ≤ G

# ) (.! "

5

# ! . .& G

A B ! |G \ (A B)| = 3 > G , # H

Z6! D6! E8! Z2 × Z4! D8! E4 × Z3! Z2 × D6

# " M E G I M =

M1 M2 . . . Mn ! M = M1 M2 . . . Mn .

# # !

M H

M Mi Mi M i {1, . . . , n}

#$ I M G

: ! M

# I G

n 4 H H1, H2, . . . , Hn! H1H2 . . . Hn

" G

# G = AB! A ≤ G, B ≤ G, a Z(A), b Z(B) > aGbG = (ab)G

# G = AB! A B E G!

A ∩B = 1! Z(G) = {a1b1, a2b2, . . . , aαbα, . . .}! aα A, bα B ( J A1 := {a1, a2, . . . , aα, . . .} B1 := {b1, b2, . . . , bα, . . .} E

G

. A1, B1 = A1 × B1 = Z(G)A1 = Z(G)B1

0 I Z(A) = Z(B) = 1! |Z(G)|2 < |G|.

# A B E # G > ( |AB| = |A||B|/|A ∩ B|D

. | A, B | = [ |A|, |B| ] · (iA, iB )! iA = | A, B : A|,

# A B E

# G >

G = AB |G| = [ |A|, |B| ].

# A B E # G ! |G| = [ |A|, |B| ] > G = AB A B G

# ! I # G 4 A B

! G = AB.

# " H K 4 : G

"

( HK G

. I : g HK l #

: H : K! # 4 #g : l #

: H : K

0 I |H| |K| H K !

HK 4 : G! 4 |HK|

|H| |K|