O Teorema de Lagrange , aplicado na teoria dos grupos , é um teorema que diz que se
G
{\displaystyle G}
grupo finito e
H
{\displaystyle H}
subgrupo de
G
{\displaystyle G}
ordem (quantidade de elementos) de
H
{\displaystyle H}
G
.
{\displaystyle G.}
Teorema de Lagrange .
Teorema 0.1
Se
⋆
{\displaystyle \star }
relação de equivalência em
S
{\displaystyle S}
S
=
⋃
[
a
]
,
{\displaystyle S=\bigcup [a],}
união é sobre um elemento de cada classe e onde
[
a
]
≠
[
b
]
{\displaystyle [a]\neq [b]}
[
a
]
∩
[
b
]
=
∅
.
{\displaystyle [a]\cap [b]=\emptyset .}
⋆
{\displaystyle \star }
S
{\displaystyle S}
classes de equivalência .
Demonstração
Seja
a
∈
S
.
{\displaystyle a\in S.}
a
∈
[
a
]
.
{\displaystyle a\in [a].}
S
=
⋃
a
∈
S
[
a
]
.
{\displaystyle S={\underset {a\in S}{\bigcup }}[a].}
Suponhamos que
[
a
]
∩
[
b
]
≠
∅
{\displaystyle [a]\cap [b]\neq \emptyset }
[
a
]
=
[
b
]
.
{\displaystyle [a]=[b].}
Seja
c
∈
[
a
]
∩
[
b
]
.
{\displaystyle c\in [a]\cap [b].}
Então
c
⋆
a
{\displaystyle c\star a}
c
⋆
b
.
{\displaystyle c\star b.}
Por um lado
{
c
⋆
a
⇔
a
⋆
c
c
⋆
b
⇒
a
⋆
b
⇒
a
∈
[
b
]
{\displaystyle {\begin{cases}c\star a\Leftrightarrow a\star c\\c\star b\end{cases}}\Rightarrow a\star b\Rightarrow a\in [b]}
Por outro
{
c
⋆
a
c
⋆
b
⇔
b
⋆
c
⇒
b
⋆
a
⇒
b
∈
[
a
]
.
{\displaystyle {\begin{cases}c\star a\\c\star b\Leftrightarrow b\star c\end{cases}}\Rightarrow b\star a\Rightarrow b\in [a].}
Seja
x
∈
[
a
]
.
{\displaystyle x\in [a].}
Então
x
⋆
a
.
{\displaystyle x\star a.}
Mas
a
⋆
b
,
{\displaystyle a\star b,}
x
⋆
b
{\displaystyle x\star b}
x
∈
[
b
]
.
{\displaystyle x\in [b].}
Portanto
[
a
]
⊂
[
b
]
.
{\displaystyle [a]\subset [b].}
y
∈
[
b
]
.
{\displaystyle y\in [b].}
Então
y
⋆
b
.
{\displaystyle y\star b.}
b
⋆
a
,
{\displaystyle b\star a,}
y
⋆
a
{\displaystyle y\star a}
y
∈
[
a
]
.
{\displaystyle y\in [a].}
Portanto
[
b
]
⊂
[
a
]
.
{\displaystyle [b]\subset [a].}
E, dessa forma,
[
a
]
=
[
b
]
.
{\displaystyle [a]=[b].}
◻
{\displaystyle \Box }
editar
Seja
⋆
{\displaystyle \star }
a
⋆
b
{\displaystyle a\star b}
a
b
−
1
∈
H
.
{\displaystyle ab^{-1}\in H.}
Temos que
[
a
]
=
H
a
=
{
h
a
|
h
∈
H
}
.
{\displaystyle [a]=Ha=\{ha~|~h\in H\}.}
Seja
k
{\displaystyle k}
G
{\displaystyle G}
H
a
1
,
…
,
H
a
k
.
{\displaystyle Ha_{1},\ldots ,Ha_{k}.}
Pelo Teorema 0.1 ,
G
=
H
a
1
∪
…
∪
H
a
k
{\displaystyle G=Ha_{1}\cup \ldots \cup Ha_{k}}
H
a
i
∩
H
a
j
=
∅
,
{\displaystyle Ha_{i}\cap Ha_{j}=\emptyset ,}
i
≠
j
.
{\displaystyle i\neq j.}
Provemos que qualquer
H
a
i
{\displaystyle Ha_{i}}
|
H
|
{\displaystyle |H|}
Seja
φ
:
H
→
H
a
i
{\displaystyle \varphi :H\to Ha_{i}}
φ
(
h
)
=
h
a
i
,
∀
h
∈
H
{\displaystyle \varphi (h)=ha_{i},~\forall h\in H}
Provemos que
φ
{\displaystyle \varphi }
Note que
φ
{\displaystyle \varphi }
φ
(
h
)
=
h
a
i
=
h
′
a
i
=
φ
(
h
′
)
{\displaystyle \varphi (h)=ha_{i}=h'a_{i}=\varphi (h')}
h
=
h
′
{\displaystyle h=h'}
H
a
i
.
{\displaystyle Ha_{i}.}
Potanto,
φ
{\displaystyle \varphi }
|
H
a
i
|
=
|
H
|
.
{\displaystyle |Ha_{i}|=|H|.}
Como
G
=
H
a
1
∪
…
∪
H
a
k
{\displaystyle G=Ha_{1}\cup \ldots \cup Ha_{k}}
H
a
i
{\displaystyle Ha_{i}}
|
H
|
{\displaystyle |H|}
|
G
|
=
k
|
H
|
.
{\displaystyle |G|=k|H|.}
Portanto,
|
H
|
{\displaystyle |H|}
|
G
|
.
{\displaystyle |G|.}
◻
{\displaystyle \Box }
![{\displaystyle S=\bigcup [a],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1be31d1c3bc5dfeee7e0e567783be7b5d1feabe3)
![{\displaystyle [a]\neq [b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01b313bd1cbb7ce96e790137dee3c1937591498e)
![{\displaystyle [a]\cap [b]=\emptyset .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4df4ff765808c7556bc7e049b62876923174983)
![{\displaystyle a\in [a].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2cca5603be2509de02ab2493e5e200575045a84)
![{\displaystyle S={\underset {a\in S}{\bigcup }}[a].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1dea0da363180a9c2d84c5b2bcdb0b01e05c4f19)
![{\displaystyle [a]\cap [b]\neq \emptyset }](https://wikimedia.org/api/rest_v1/media/math/render/svg/44cc5708a5f3d11a4d51a53bd969c0de6f3cc57f)
![{\displaystyle [a]=[b].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cea6f461568f4e243b1bdc9d44a7ee9d63aa8da)
![{\displaystyle c\in [a]\cap [b].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7badb7905f72e7142db195fd6b5af9f2352b050)
![{\displaystyle {\begin{cases}c\star a\Leftrightarrow a\star c\\c\star b\end{cases}}\Rightarrow a\star b\Rightarrow a\in [b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0da53565c43581f0305d2f9ddaa0537f1a8ccb10)
![{\displaystyle {\begin{cases}c\star a\\c\star b\Leftrightarrow b\star c\end{cases}}\Rightarrow b\star a\Rightarrow b\in [a].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb2f792c276264c979ab315dbb7651af2447e035)
![{\displaystyle x\in [a].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41886156bcd0fd1b29bf3100cabe31455a72e31d)
![{\displaystyle x\in [b].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec319200d2a40e1a1371448281c1e6ffc07602e5)
![{\displaystyle [a]\subset [b].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e23b9209abb4f2e583f169560a79cb3bdd512c56)
![{\displaystyle y\in [b].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/874a5626f7e17ac34789cff12d6beee8c9c084c5)
![{\displaystyle y\in [a].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65dbc1a17f35e7f05dcb01adecd1d711221682e3)
![{\displaystyle [b]\subset [a].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/334e52a1580ea9a6ad56e6d80a2fae99774975f8)
![{\displaystyle [a]=Ha=\{ha~|~h\in H\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab2b37b4678e692a2df3f6640029b9aa6257fa9b)
