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R. HAZRAT AND A. R. WADSWORTH

T h e o r e m 3 . 4 . L e t E b e a n y g r a d e d d i v i s i o n r i n g n i t e d i m e n s i o n a l o v e r i t s c e n t e r T . S o , Z ( E 0 ) i s G a l o i s

o v e r T 0 ; l e t G = a l ( Z ( E 0 ) / T 0 ) . L e t δ = i n d ( E ) i n d ( E 0 ) [ Z ( E 0 ) : T 0 ] , a n d l e t µ δ ( T 0 ) b e t h e g r o u p o f t h o s e δ - t h r o o t s o f u n i t y l y i n g i n T 0 . L e e t N = N Z ( E 0 ) / T 0 N r d E 0 : E 0 T 0 . T h e n , t h e r o w s a n d c o l u m n o f

the following diagram are exact:

1

S K 1 ( E 0 )

//

e k e r N / [ E 0 , E ]

### NrdE0

//

b H

1 ( G , N r d E 0 ( E 0 ) )

//

1

## T∧

E

T

//

E ( 1 ) / [ E 0 , E ]

//

SK1(E)

//

1

(3.5)

e N

e µ δ ( T 0 ) N ( E 0 )

1

The (γ +

map

in (3.5) is given as follows: For γ, δ

# T ) ∧ (δ +

T ) = [ x γ , x δ ] m o d [ E 0

, E].

E , t a k e a n y n o n z e r o x γ E γ a n d x δ E δ . T h e n ,

P r o o f . B y P r o p . 2 . 3 i n [ H w W 2 ] , Z ( E 0 ) / T 0 i s a G a l o i s e x t e n s i o n a n d t h e m a p θ : E A u t ( E 0 ) , g i v e n b y e 7 ( a 7 e a e 1 ) f o r a E 0 , i n d u c e s a n e p i m o r p h i s m E G = a l ( Z ( E 0 ) / T 0 ) . I n t h e n o t a t i o n f o r ( 3 . 4 )

w i t h A = N r d E 0 ( E 0

),

we

have

N G

c o i n c i d e s w i t h N Z ( E 0 ) / T 0 k e r ( N G ) = N r d E e 0 ( k e r ( N ) ) . on A. Hence,

(3.6)

T a k e a n y e E a n d l e t σ = θ ( e ) A u t ( E 0 ) . F o r a n y a E 0 , l e t h a Z ( T 0 ) [ x ] b e t h e m i n i m a l p o l y n o m i a l o f a o v e r Z ( T 0 ) . T h e n σ ( h a ) Z ( T 0 ) [ x ] i s t h e m i n i m a l p o l y n o m i a l o f σ ( a ) o v e r Z ( T 0 ) . H e n c e , N r d E 0 ( σ ( a ) ) = σ ( N r d E 0 ( a ) ) . S i n c e σ | Z ( T 0 ) N r d E 0 ( [ a , e ] ) G, this yields = N r d E 0 ( a σ ( a 1 ) ) = N r d e 1 E 0 ( a ) σ ( N r d E 0 ( a IG(A), (3.7) ) ) h e n e c e N ( [ a , e ] ) = 1 . T h u s , w e h a v e [ E 0 , E ] k e r ( N ) E ( with the latter inclusion from Prop. 3.2(iv). 1 )

T h e f o r m u l a i n P r o p . 3 . 2 ( i v ) a l s o s h o w s t h a t e N ( E ( 1 ) ) µ δ ( T 0 ) . T h u s , t h e v e r t i c a l m a p s i n d i a g r a m ( 3 . 5 )

e a r e w e l l - d e fi n e d , a n d t h e c o l u m n i n ( 3 . 5 ) i s e x a c t . B e c a u s e N r d E 0 m a p s k e r ( N ) o n t o k e r ( N G ) b y ( 3 . 6 ) a n d

i t m a p s [ E 0 , E ] o n t o I G ( A ) b y ( 3 . 7 ) ( a s θ ( E ) m a p s o n t o G ) , t h e m a p l a b e l l e d N r d E 0

in diagram (3.5)

i s s u r j e c t i v e w i t h k e r n e l E ( 1 ) 0

r o w , s i n c e

[ E , E ]

E 0

an

[ E 0 , E ] [ E 0 , E ] . T h = E/ e r e f o r e , t h e t o p r o w o f ( 3 . 5 ) i s e x a c t . F o r t h e l o w e r d E ( E 0 Z ( E ) ) T , the following lemma yields an epimorphism

E / T E / T [ E , E ] / [ E 0 , E ] . G i v e n t h i s , t h e l o w e r r o w i n ( 3 . 5 ) i s e v i d e n t l y e x a c t .

Lemma 3.5. Let G be a group, and let H be a subgroup of G with H [G, G]. Let B = G(H Z(G)). Then, there is an epimorphism B B [G, G][H, G].

 Proof. Since [G, G] H, we have [G, G], [G, G] [H, G], so [G, H] is a normal subgroup of [G, G] with

abelian factor group. Consider the map β : G × G [G, G]/[H, G] given by (a, b) 7→ aba

1b

1[H, G]. For

any a, b, c G we have the commutator identity [a, bc] = [a, b] [b, [a, c]] [a, c]. The middle term [b, [a, c]] lies in [H, G]. Thus, β is multiplicative in the second variable; likewise, it is multiplicative in the first variable. A s [ H Z ( G ) , G ] [ H , G ] , t h i s β i n d u c e s a w e l l - d e fi n e d g r o u p h o m o m o r p h i s m β 0 : B B [ G , G ] / [ H , G ] , w h i c h i s s u r j e c t i v e s i n c e i m ( β ) g e n e r a t e s [ G , G ] / [ H , G ] . S i n c e β 0 ( η η ) = 1 f o r a l l η B , t h e r e i s induced epimorphism B B [G, G]/[H, G]. a n

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