Niech
U
V
^
:=
U
∩
V
.
{\displaystyle {\widehat {UV}}:=U\cap V.}
B
⊴
A
,
{\displaystyle B\trianglelefteq A,}
[a]
B
∩
C
⊴
A
∩
C
,
{\displaystyle B\cap C\trianglelefteq A\cap C,}
B
C
^
⊴
A
C
^
;
{\displaystyle {\widehat {BC}}\trianglelefteq {\widehat {AC}};}
C
⊴
D
{\displaystyle C\trianglelefteq D}
A
D
^
⊴
A
C
^
.
{\displaystyle {\widehat {AD}}\trianglelefteq {\widehat {AC}}.}
B
C
^
⊴
A
C
^
{\displaystyle {\widehat {BC}}\trianglelefteq {\widehat {AC}}}
A
D
^
⊴
A
C
^
,
{\displaystyle {\widehat {AD}}\trianglelefteq {\widehat {AC}},}
E
=
B
C
^
A
D
^
=
A
D
^
B
C
^
,
{\displaystyle E={\widehat {BC}}\,{\widehat {AD}}={\widehat {AD}}\,{\widehat {BC}},}
E
⊴
A
C
^
{\displaystyle E\trianglelefteq {\widehat {AC}}}
iloczyn kompleksowy ).
Ponieważ
E
⊴
A
C
^
⩽
A
{\displaystyle E\trianglelefteq {\widehat {AC}}\leqslant A}
B
⊴
A
,
{\displaystyle B\trianglelefteq A,}
[b]
B
E
⊴
B
A
C
^
oraz
B
A
C
^
/
B
E
≃
A
C
^
/
E
(
B
∩
A
C
^
)
.
(
1
)
{\displaystyle BE\trianglelefteq B{\widehat {AC}}\quad {\text{oraz}}\quad B{\widehat {AC}}/BE\simeq {\widehat {AC}}/E(B\cap {\widehat {AC}}).\qquad (1)}
Teraz
B
E
=
B
(
B
C
^
A
D
^
)
=
(
B
B
C
^
)
A
D
^
=
B
A
D
^
{\displaystyle BE=B({\widehat {BC}}\,{\widehat {AD}})=(B{\widehat {BC}}){\widehat {AD}}=B{\widehat {AD}}}
E
(
B
∩
A
C
^
)
=
E
{\displaystyle E(B\cap {\widehat {AC}})=E}
B
∩
A
C
^
=
B
C
^
⊆
E
{\displaystyle B\cap {\widehat {AC}}={\widehat {BC}}\subseteq E}
B
A
D
^
⊴
B
A
C
^
oraz
B
A
C
^
/
B
A
D
^
≃
A
C
^
/
E
.
(
2
)
{\displaystyle B{\widehat {AD}}\trianglelefteq B{\widehat {AC}}\quad {\text{oraz}}\quad B{\widehat {AC}}/B{\widehat {AD}}\simeq {\widehat {AC}}/E.\qquad (2)}
Powtarzając to samo rozumowanie dla
A
,
B
{\displaystyle A,B}
C
,
D
,
{\displaystyle C,D,}
D
B
C
^
⊴
D
A
C
^
oraz
D
A
C
^
/
D
B
C
^
≃
A
C
^
/
E
.
(
3
)
{\displaystyle D{\widehat {BC}}\trianglelefteq D{\widehat {AC}}\quad {\text{oraz}}\quad D{\widehat {AC}}/D{\widehat {BC}}\simeq {\widehat {AC}}/E.\qquad (3)}
Teza wynika z połączenia (2) oraz (3).
↑ Lemat 1 . Niech
B
⊴
A
⩽
G
{\displaystyle B\trianglelefteq A\leqslant G}
C
⩽
G
.
{\displaystyle C\leqslant G.}
B
∩
C
⊴
A
∩
C
{\displaystyle B\cap C\trianglelefteq A\cap C}
A
∩
C
/
B
∩
C
≃
B
(
C
∩
A
)
/
B
.
{\displaystyle A\cap C/B\cap C\simeq B(C\cap A)/B.}
Dowód . Drugie twierdzenie o izomorfizmie mówi, że jeśli
G
{\displaystyle G}
H
⊴
G
{\displaystyle H\trianglelefteq G}
K
⩽
G
,
{\displaystyle K\leqslant G,}
H
∩
K
⊴
K
{\displaystyle H\cap K\trianglelefteq K}
K
/
H
∩
K
{\displaystyle K/H\cap K}
H
K
/
H
.
{\displaystyle HK/H.}
G
,
H
,
K
{\displaystyle G,H,K}
A
,
B
,
A
∩
C
,
{\displaystyle A,B,A\cap C,}
B
∩
(
A
∩
C
)
⊴
A
∩
C
{\displaystyle B\cap (A\cap C)\trianglelefteq A\cap C}
A
∩
C
/
B
∩
(
A
∩
C
)
≃
B
(
A
∩
C
)
/
B
.
{\displaystyle A\cap C/B\cap (A\cap C)\simeq B(A\cap C)/B.}
B
∩
(
A
∩
C
)
=
B
∩
C
,
{\displaystyle B\cap (A\cap C)=B\cap C,}
B
(
A
∩
C
)
=
B
(
C
∩
A
)
,
{\displaystyle B(A\cap C)=B(C\cap A),}
↑ Lemat 2 . Niech
A
⊴
C
⩽
G
{\displaystyle A\trianglelefteq C\leqslant G}
B
⊴
G
.
{\displaystyle B\trianglelefteq G.}
B
A
⊴
B
C
{\displaystyle BA\trianglelefteq BC}
B
C
/
B
A
≃
C
/
A
(
B
∩
C
)
.
{\displaystyle BC/BA\simeq C/A(B\cap C).}
Dowód . Ponieważ
B
⊴
G
,
{\displaystyle B\trianglelefteq G,}
A
B
=
B
A
⩽
G
{\displaystyle AB=BA\leqslant G}
C
B
=
B
C
⩽
G
.
{\displaystyle CB=BC\leqslant G.}
B
A
⩽
B
C
.
{\displaystyle BA\leqslant BC.}
B
A
{\displaystyle BA}
B
C
.
{\displaystyle BC.}
B
⩽
B
A
⩽
N
G
(
B
A
)
{\displaystyle B\leqslant BA\leqslant N_{G}(BA)}
normalizator ), otrzymuje się
(
B
A
)
b
=
B
A
{\displaystyle (BA)^{b}=BA}
b
∈
B
{\displaystyle b\in B}
sprzężenie i podgrupa normalna ). Wtedy, dla dowolnych
b
∈
B
,
c
∈
C
,
{\displaystyle b\in B,\,c\in C,}
(
B
A
)
b
c
=
[
(
B
A
)
b
]
c
=
(
B
A
)
c
=
B
c
A
c
=
B
A
c
=
B
A
,
{\displaystyle (BA)^{bc}=[(BA)^{b}]^{c}=(BA)^{c}=B^{c}A^{c}=BA^{c}=BA,}
B
◃
G
{\displaystyle B\triangleleft G}
A
⊴
C
.
{\displaystyle A\trianglelefteq C.}
(
B
A
)
x
=
B
A
{\displaystyle (BA)^{x}=BA}
x
∈
B
C
{\displaystyle x\in BC}
B
A
⊴
B
C
.
{\displaystyle BA\trianglelefteq BC.}
drugiego twierdzenia o izomorfizmie z
B
C
,
B
A
,
C
{\displaystyle BC,BA,C}
G
,
H
,
K
{\displaystyle G,H,K}
A
B
∩
C
⊴
C
{\displaystyle AB\cap C\trianglelefteq C}
C
/
A
B
∩
C
≃
C
(
A
B
)
/
A
B
.
{\displaystyle C/AB\cap C\simeq C(AB)/AB.}
A
B
∩
C
=
A
(
B
∩
C
)
{\displaystyle AB\cap C=A(B\cap C)}
C
(
A
B
)
=
(
C
A
)
B
=
C
B
=
B
C
,
{\displaystyle C(AB)=(CA)B=CB=BC,}
C
/
A
(
B
∩
C
)
≃
B
C
/
B
A
.
{\displaystyle C/A(B\cap C)\simeq BC/BA.}
↑ Zob. Pierce, s. 27, ćw. 1.
↑ The Butterfly and the Serpent. W: J. Lambek: Logic and Algebra . Aldo Ursini, Paulo Agliano (red.). CRC Press, 1996, s. 161–180. ISBN 978-0-8247-9606-8 . Bibliografia
edytuj
![{\displaystyle (BA)^{bc}=[(BA)^{b}]^{c}=(BA)^{c}=B^{c}A^{c}=BA^{c}=BA,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99042476c7642f496a162301456956863965b8c1)
