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Белоногов. Задачник по теории групп

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% H - 7 , H 1! Z2! E4! D8! P

& H 3, 1, 2

! 0 H a × b2 ! a2 × b ! a2 × ab

! Φ(P ) = Ω1(P ) = ap × bp

! " P p+1 H abi i {1, . . . p−

1} ap × b

pn−1−i

× b

H Ωi(P ) = a

 

!$

(( H 2, 2

 

!

3 ( (-./ (. H 2, 2

!

Zpn n ≥ 0! Ep2! Q8

!

Zpn × Zp n ≥ 2! Mpn pn = 8

! !

D8! Q8! Mpn (n ≥ 3)

!$

1(P ) Z(P ) m > 1 n > 1

!

7 P = Q8 Z4 - H 0

! , D8! 0 ! , Z4 × Z2!

! , " Q8 m(P ) = 2! rs(P ) = 3

! !

7 P = D8 D8 (/ !

, D8 Z4 ( Q8 Z4)! & ! ,

Q8 × Z2D m(P ) = 2! rs(P ) = 4

7 P = D8 Q8 5 ! ,

Q8 Z4 8 ! , D8 × Z2D m(P ) = 3!

rs(P ) = 4

 

 

 

 

§18

" "

% H I C E ! p = 2 C = Aut(P )!

p > 2 C =

α ! α E , pn−1 #

xα = x1+p

 

 

 

 

"

 

D8

 

" #

 

( H < P Zp

 

 

 

§19

# "

g E

#

 

H -.! 05

# 0 H G 0 H & .

.%! (/ 0

(.! 5 & ./

# ! Diag2(q)! U T2(q) × Z(G)

# # . H Z "$# q2 1 4 q q2 +q+1

4 q

#

Z p pm + 1

# !

. H U T3(q) D! D Zq−1 × Zq−1 0 H {t31(a) | a Fq}

#$

CG(H) = Diag3(q)! CG(H) ∩ SL3(q) Zq−1 × Zq−1

#

GLn−1(q)

#

F +

#

+ # (8 &( (

§20

$ . .! .! .! ( H

4 E . D DE

.

$ l2(G) 2

$ . H Oπ(G) × Oπ (G)$! . H G = A4 × S3

§21

; D8

2 H Z2 × A5

. H G = S4! B = a 4 b 2 Syl2(G)! A = a ! H = A4

. H G = S4! B = a 4 b 2 Syl2(G)! A = {a, a1}

 

 

1

1

1

 

 

 

§22

 

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# ....

 

 

 

 

 

 

 

 

 

(χ, χ) = χ(1)2

χ(1) = 1

 

 

 

 

 

; $ S3 ...6

 

( ,$" ωχ

..%( ) χ(1)χ(gm)

# χ Irr(G)! ...%

$ . H ; ψ Irr(G)! Irr(χ)ψ Irr(G)

|Irr(χ)ψ| = Irr(χ)

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" A

, ! ,

! ..00 (% ./

k

! aχ = i=1 χ(gi)! g1, . . . , gk E

4 : G 3 : " #

$ E $ $

"$ Ker (AGT ) = Ker (A)G

§23

+ # 0 (5

+ # .0 ( .0 .

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M = a × Q1! Q1 ≤ Q! 4 M E

G a 3 ! Q1

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! Φ(Q) = 1 > Ω(Q) (Q) Φ(Q) Z(G) !

#
$
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t := [g, a] q ! ! t Z(G) 2 ga = gt! ga2 = gt2, . . . , g = gap = gtp! tp = 1!

+ # † Wh zQi* .5% mh`*~dh*` .0j .( .- = G E ! G Zp ×S! S E p

G p! E p

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" G = Z(G)S! S E

@ pn pq @

G = D18! N Z3

H @ H H 2

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a p Q

Q! #

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#

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# H G , Z3 S3 H G , Z4! E4! A4

D10

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9 #

§1

$ 6

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D(A) (8

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SLn(C) ./

./

§2

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a .(

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G : H .(

|G : H| .(

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= .%

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Q8 .5

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$ p 0/

p0/

§3

"

0.

CG(g) 0.

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π 0. p: 0. p 0.

# : 0.

p 0. : p 0.

4 : 00 D2n 08

: 08

D08

:

08

p # Q(p) %.

:

%. o(a; H) %.

§4

%%

H G %%

%%

%%

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NG(M) := {g G | Mg = gM}

%%

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56

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§9

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((.

§10

p ((0

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p

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π ((0 p ((0

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3 ((&

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§11

: , (.0

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(.0

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(.0

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(.-

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§12

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# (0.

[A, B] (0.

4 (00

§13

(08

# (08

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(08

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§14

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