Белоногов. Задачник по теории групп
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6 T := a1 × a2 E4 T P
8 T1 := z × c E4 P !
T ! T1
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a c P ! (aic)P = (aic) a = P aic |
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(0 Q := P d Q2n+1! Q = dP P Q E
P ! , Q2n+1
(% Ω1(P ) = T Q P/Ω1(P ) Z2n−1
(& P = Q ac ! (ac)2 = ab
(5 M2 = Z(P )Q Z2n Q2n+1
(- I Q < H < P ! H = Z(H)Q T ≤ H
(6 I H E P ! Z(H) ≤ Z(P )
(8 I X A ≤ P ! X E $ P A E $ 2n P ! P = X A
./ m(P ) = 2 rs(P ) = 3
! P E D8 × Z2! Q8 × Z2! D8 Z4 2 #|P | exp(P )!
Z(P )! P ! Φ(P )!
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! P {D8 × D8, D8 × Q8, Q8 × Q8} 2 # Z(P )! P ! Φ(P )!
: P ! Ω1(P )! d(P )! m(P ) rs(P ) 3 ! , E2m(P )! P O
! |P | = 16 m(P ) = 3 > P E
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! P = E x ! E E8 o(x) = 2 > |CP (x)| ≥ 4
! P = E x ! E E2n (n ≥ 1) o(x) = 2 >
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( ) ϕ : α → α|H (α Aut(G)) ,
Aut(G) Aut(H) Ker(ϕ) = CAut(G)(H) 3 !
Aut(G)/CAut(G)(H) ≤ Aut(H).
. ) ψ : α → αˆ (α Aut(G))! αˆ : gH → gαH
g G! , Aut(G) → Aut(G/H) Ker(ψ) =
CAut(G)(G/H) (= {α Aut(G) | gα gH g G} ) 3
!
Aut(G)/CAut(G)(G/H) ≤ Aut(G/H).
" > P E pn d = d(P ) ( 3 ,
γ: Aut(P ) → Aut(P/Φ(P )) GLd(Zp),
p !
γ: α → αˆ (α Aut(P )), αˆ : gΦ(P ) → gαΦ(P ),
! ! Ker(γ) = CAut (P )(P/Φ(P ))
. |CAut (P )(P/Φ(P ))| pd(n−d)
|Aut (P )| pd(n−d)(pd − 1)(pd − p) . . . (pd − pd−1).
" > A E p Aut(P )
( A ≤ Aut(P/Φ(P ))
. P = [P, A]CP (A) [P, A] = [[P, A], A]
0 I P ! P = [P, A] × CP (A)
% I A $ # # P !
A = 1
" G = a pn b m! p m > b G!
CG(b) = b
" > I P E ! A E p Aut(P ) A
$ Ω1(P )! A = 1 |
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, α! xα = x1+p
0 I p = 2 n |
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2! Aut (P ) = α |
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xβ = x−1 |
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% CAut (P )(P/Φ(P ))O
" # Hol(Zpn ) Z+pn ϕ Z·pn ! xϕ(y) = xy x Z+pn y Z·pn
"$ I G = a 169 ( b 15! G = ( a b5 ) × b3
" I P Z2m × Z2n m > n! Aut (P ) E .
" Aut(Z4 × Z2) O
" P = a 2m × b 2 m > 2 > |Aut (P )| = 2m+1
Aut (P ) = (( α 2m−2 × β 2) γ 2) × δ 2,
p
(a, b)α = (a5, b)! (a, b)β = (ab, b)! (a, b)γ = (a, a2m−1 b)! (a, b)δ = (a−1, b)
:
αγ = α, βγ = α2m−3 β, α × β × δ = CAut(P )(Ω1(P )).
" I P Zpm × . . . × Zpm ! Aut (P ) GLn(Zpm ) 5 ./ |
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" P = a1 pm × . . . × an pm |
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( P = a1x1 × . . . × anxn |
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P = b1 × . . . × bn ! |GLn(p)| · |Φ(P )|n |
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P = a 4 × b 4 A = Aut (P ) |
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( ) ϕ (a, b)ϕ = (ab, ab−1)
, P
. |A| = 25 · 3
0 A " .
T:= ( α 2 × β 2) ( γ 4 δ 2),
(a, b)α = (a−1, b)! (a, b)β = (a, b−1)! (a, b)γ = (b, ab2)! (a, b)δ = (b, a)
: γ δ D8! αγ = αδ = β! βγ = βδ = α
% O2(A) = α×β×γ2×γδ = CAut(P )(P/Φ(P )) = CAut(P )(Φ(P ))
& T E4 Z2
5 A = O2(A) S! S S3
" ! ( Aut(D2n ) E . n ≥ 3! ! Aut(D2n )
Hol(Z2n )
. Aut(D8) D8
" " ( Aut(Q2n ) E . n > 3!
. Aut(Q8) S4
" # P D8 × Z2
( |Aut(P )| = 26 = 64
. Out(P ) D8 × Z2
0 A Aut(P ) E! , E16
CAut(P )(P/Φ(P )) E8
"$ P Q8 × Z2
( |Aut(P )| = 26 · 3
. 3 . Aut(P ) E F ! E E16
F E4
" P Mpn > Aut(P )
! ! Aut(P )/Op(Aut(P )) ≤ Zp−1 × Zp−1 A ! Aut(M2n ) E .
" P E(p)! P = ( a p × z p) b p! p > 2! z Z(P )
ab = az > Aut(P ) Hol(Ep2 )
" G = E ( x n τ 2)! E E2m ! x τ D2n M
4 n x # E E x E
9 E > |CE (τ )| = |E|1/2 "$ E τ \ E τ E τ
" G = P A! P Sylp(G) B ≤ P I A $
B CP (B)! A $ P
" A×B > A×B ≤ Aut(P )! A E p !
B E p I A $ CP (B)! A = 1
" P, A, B E # G ! P G! P B E p! A = Op(A) [A, B] = 1 > [Cp(B), A] = 1![P, A] = 1
" ! B ≤ P G! P E p Cp(B) B > Op(CG(B))
CG(P )
" " G = P R! P Sylp(G) I R $
P/Z(P ) [P, R] Z(P )! P = [P, R] × CP (R)
" # G = T S! |S| = 3 T , #
Z4! E4! Q8! D8 > G = T × S! T = [T, S]
"$ Hol (Q8) = Q (E S)! Q Q8! E E4! S S3! E S S4Q E Q8 Q8
" G = Q R! Q Syl2(G) Q Q2n (n ≥ 3) >
CR(Q) = 1 R # Q! R = 1!
G SL(2, 3)
" P D8 D8 ( Q8 Q8)
( P ! , Q8
. |Aut (P )| = 27 · 3 Aut (P )/Inn (P ) S3 Z2
0 B E 0 Aut(P )! # "
P/Z(P ) > " "$
P ! B!
" P D8 Q8
( P ! , # E8
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0 3
P " Z5
" P = H1 × . . . × Hn! Hi D2m (n, m N, m ≥ 3)
Z(Hi) = zi
( # : Aut (P ) $
{z1, . . . , zn}
. {α Aut (P ) | ziα = zi i {1, . . . , n}} . Aut (P ) " > p > 2 A E p Aut(P ) I A $
Ω1(P )! A = 1
" A GL2(Zp) ! , E8
" ! G = P E! P E p! p > 2! E E2n ! n ≥ 3 I CE (P ) = 1! E # : SCN3(P )
" " p > 2 P = a, b, c ap2 = bp3 = cp4 = 1, ba = b1+p2, cb = c1+p3 , ac = a1+p >