Definisi dasasr
sunting
Definisi:
Misalkan
G
{\displaystyle G}
Operasi biner pada
G
{\displaystyle G}
∘
:
G
×
G
→
G
{\displaystyle \circ :G\times G\rightarrow G}
a
∘
b
∈
G
{\displaystyle a\circ b\in G}
a
,
b
∈
G
{\displaystyle a,b\in G}
Suatu himpuan
G
{\displaystyle G}
∘
{\displaystyle \circ }
grupoid . Notasi:
(
G
,
∘
)
{\displaystyle (G,\circ )}
Contoh:
(
N
,
+
)
,
(
Z
,
+
)
,
(
N
,
⋅
)
{\displaystyle (\mathbb {N} ,+),(\mathbb {Z} ,+),(\mathbb {N} ,\cdot )}
(
Z
,
⋅
)
{\displaystyle (\mathbb {Z} ,\cdot )}
(
K
,
+
)
{\displaystyle (\mathbb {K} ,+)}
(
K
,
⋅
)
{\displaystyle (\mathbb {K} ,\cdot )}
K
=
Q
,
R
{\displaystyle \mathbb {K} =\mathbb {Q} ,\mathbb {R} }
C
{\displaystyle \mathbb {C} }
Definisi :
Suatu grupoid
(
G
,
∘
)
{\displaystyle (G,\circ )}
grup jika ia memenuhi hukum-hukum berikut:
(G1)
(
a
∘
b
)
∘
c
=
a
∘
(
b
∘
c
)
{\displaystyle (a\circ b)\circ c=a\circ (b\circ c)}
a
,
b
,
c
∈
G
{\displaystyle a,b,c\in G}
Hukum assotiatif )(G2) Terdapat suatu anggota
e
∈
G
{\displaystyle e\in G}
a
∘
e
=
e
∘
a
=
a
{\displaystyle a\circ e=e\circ a=a}
a
∈
G
{\displaystyle a\in G}
e
{\displaystyle e}
anggota identitas )(G3) Terdapat suatu anggota identitas
e
∈
G
{\displaystyle e\in G}
a
∈
G
{\displaystyle a\in G}
b
∈
G
{\displaystyle b\in G}
a
∘
b
=
b
∘
a
=
e
{\displaystyle a\circ b=b\circ a=e}
Hukum invers , Notasi
b
=
a
−
1
{\displaystyle b=a^{-1}}
Suatu grup
(
G
,
∘
)
{\displaystyle (G,\circ )}
grup Abelian (atua grup komutatif ) jika itu memnuhi jugaasi
(G4)
a
∘
b
=
b
∘
a
{\displaystyle a\circ b=b\circ a}
a
,
b
∈
G
{\displaystyle a,b\in G}
Hukum komutatif )Untuk mendefinisikan grup kita dapat menganti (G2) dan (G3) dengan (G2)' dan (G3)' berikut:
(G2)' Terdapat suatu anggota
e
∈
G
{\displaystyle e\in G}
a
∘
e
=
a
{\displaystyle a\circ e=a}
a
∈
G
{\displaystyle a\in G}
e
{\displaystyle e}
anggota identitas kanan )(G3)' Terdapat suatu anggota identitas kanan
e
∈
G
{\displaystyle e\in G}
a
∈
G
{\displaystyle a\in G}
b
∈
G
{\displaystyle b\in G}
a
∘
b
=
e
{\displaystyle a\circ b=e}
Hukum invers kanan )Kita membuktikan fakta di atas sebagai Lemma 1, Lemma 2 dan Lemma 3.
Lemma 1:
Misalkan
(
G
,
∘
)
{\displaystyle (G,\circ )}
(G1) , (G2)' dan (G3)' dengan anggota identitas kanan
e
∈
G
{\displaystyle e\in G}
(G3)' . Jika
a
,
b
∈
G
{\displaystyle a,b\in G}
a
∘
b
=
e
{\displaystyle a\circ b=e}
b
∘
a
=
e
{\displaystyle b\circ a=e}
Bukti:
Dengan (G3)' terdapat
c
∈
G
{\displaystyle c\in G}
b
∘
c
=
e
{\displaystyle b\circ c=e}
(G1) dan (G2)'
e
=
b
∘
c
=
(
b
∘
e
)
∘
c
=
(
b
∘
(
a
∘
b
)
)
∘
c
=
(
(
b
∘
a
)
∘
b
)
∘
c
=
(
b
∘
a
)
∘
(
b
∘
c
)
=
(
b
∘
a
)
∘
e
=
b
∘
a
{\displaystyle e=b\circ c=(b\circ e)\circ c=(b\circ (a\circ b))\circ c=((b\circ a)\circ b)\circ c=(b\circ a)\circ (b\circ c)=(b\circ a)\circ e=b\circ a}
Lemma 2:
Misalkan
(
G
,
∘
)
{\displaystyle (G,\circ )}
(G1) , (G2)' dan (G3)' dengan anggota identitas kanan
e
∈
G
{\displaystyle e\in G}
(G3)' . Maka
e
∘
a
=
a
{\displaystyle e\circ a=a}
a
∈
G
{\displaystyle a\in G}
Bukti:
Misalkan
a
∈
G
{\displaystyle a\in G}
(G3)' terdapat
b
∈
G
{\displaystyle b\in G}
a
∘
b
=
e
{\displaystyle a\circ b=e}
b
∘
a
=
e
{\displaystyle b\circ a=e}
(G1) dan (G2)'
e
∘
a
=
(
b
∘
a
)
∘
a
=
b
∘
(
a
∘
b
)
=
b
∘
e
=
b
.
{\displaystyle e\circ a=(b\circ a)\circ a=b\circ (a\circ b)=b\circ e=b.}
Lemma 3:
Misalkan
(
G
,
∘
)
{\displaystyle (G,\circ )}
(G1) , (G2)' dan (G3)' . Jika terdapat hanya satu anggota identitas.
Bukti:
Misalkan
e
∈
G
{\displaystyle e\in G}
(G3)' dan
e
′
∈
G
{\displaystyle e'\in G}
a
∘
e
′
=
e
′
{\displaystyle a\circ e'=e'}
a
∈
G
{\displaystyle a\in G}
(G2)' maka
e
∘
e
′
=
e
′
{\displaystyle e\circ e'=e'}
a
=
e
′
{\displaystyle a=e'}
e
∘
e
′
=
e
{\displaystyle e\circ e'=e}
e
′
=
e
{\displaystyle e'=e}
Teorema (Sifat kanselasi):
Misalkan
a
,
b
,
c
∈
G
{\displaystyle a,b,c\in G}
Jika
a
∘
b
=
a
∘
c
{\displaystyle a\circ b=a\circ c}
b
=
c
{\displaystyle b=c}
Jika
b
∘
a
=
c
∘
a
{\displaystyle b\circ a=c\circ a}
b
=
c
{\displaystyle b=c}
Bukti :
Dengan (G3) terdapat
d
∈
G
{\displaystyle d\in G}
a
∘
d
=
∘
d
∘
a
=
e
{\displaystyle a\circ d=\circ d\circ a=e}
a
∘
b
=
a
∘
c
{\displaystyle a\circ b=a\circ c}
(G1)
b
=
e
∘
b
=
(
d
∘
a
)
∘
b
=
d
∘
(
a
∘
b
)
=
d
∘
(
a
∘
c
)
=
(
d
∘
a
)
∘
c
=
e
∘
c
=
c
.
{\displaystyle b=e\circ b=(d\circ a)\circ b=d\circ (a\circ b)=d\circ (a\circ c)=(d\circ a)\circ c=e\circ c=c.}
Jika
b
∘
a
=
c
∘
a
{\displaystyle b\circ a=c\circ a}
(G1) dan (G2)
b
=
b
∘
e
=
b
∘
(
a
∘
d
)
=
(
b
∘
a
)
∘
d
=
(
c
∘
a
)
∘
d
=
c
∘
(
a
∘
d
)
=
c
∘
e
=
c
.
{\displaystyle b=b\circ e=b\circ (a\circ d)=(b\circ a)\circ d=(c\circ a)\circ d=c\circ (a\circ d)=c\circ e=c.}
Teorema
(
a
−
1
)
−
1
=
a
{\displaystyle (a^{-1})^{-1}=a}
Bukti:
a
−
1
∘
a
=
a
∘
a
−
1
{\displaystyle a^{-1}\circ a=a\circ a^{-1}}
a
−
1
{\displaystyle a^{-1}}
a
{\displaystyle a}
a
−
1
{\displaystyle a^{-1}}
Definisi:
Misalkan
(
G
,
∘
)
{\displaystyle (G,\circ )}
H
⊆
G
{\displaystyle H\subseteq G}
grup bagian (persisnya
(
H
,
∘
)
{\displaystyle (H,\circ )}
(
G
,
∘
)
{\displaystyle (G,\circ )}
e
∈
H
{\displaystyle e\in H}
Jika
a
,
b
∈
H
{\displaystyle a,b\in H}
a
∘
b
∈
H
{\displaystyle a\circ b\in H}
Jika
a
∈
H
{\displaystyle a\in H}
a
−
1
∈
H
{\displaystyle a^{-1}\in H}
Teorema:
Misalkan
(
G
,
∘
)
{\displaystyle (G,\circ )}
H
⊆
G
{\displaystyle H\subseteq G}
(
H
,
∘
)
{\displaystyle (H,\circ )}
a
∘
b
−
1
∈
H
{\displaystyle a\circ b^{-1}\in H}
a
,
b
∈
H
{\displaystyle a,b\in H}
Bukti:
Misalkan
(
H
,
∘
)
{\displaystyle (H,\circ )}
a
,
b
∈
H
{\displaystyle a,b\in H}
3 Definisi di atas
b
−
1
∈
H
{\displaystyle b^{-1}\in H}
2 juga
a
∘
b
−
1
∈
H
{\displaystyle a\circ b^{-1}\in H}
a
∘
b
−
1
∈
H
{\displaystyle a\circ b^{-1}\in H}
a
,
b
∈
H
{\displaystyle a,b\in H}
1 , 2 dan 3 . Karena
H
{\displaystyle H}
a
∈
H
{\displaystyle a\in H}
1 benar karena
e
=
a
∘
a
−
1
∈
H
{\displaystyle e=a\circ a^{-1}\in H}
3 benar karena
a
−
1
=
e
∘
a
−
1
∈
H
{\displaystyle a^{-1}=e\circ a^{-1}\in H}
2 juga benar karena
a
∘
b
=
a
∘
(
b
−
1
)
−
1
∈
H
{\displaystyle a\circ b=a\circ (b^{-1})^{-1}\in H}
