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Teorema-Teorema atau Sifat-sifat Grup


Setelah memahami Definisi Grup dan Cara Membuktikan Suatu Himpunan Beserta Operasinya adalah Grup atau tidak, sekarang marilah perhatikan teorema-teorema berikut. (Catatan: a*b selanjutnya cukup ditulis ab).

Teorema 1: Unsur identitas pada suatu grup bersifat tunggal.

BUKTI:  Untuk membuktikan bahwa unsur identitas grup tunggal, kita dapat menunjukkan bahwa jika dua objek memenuhi sifat identitas maka dua objek tersebut haruslah sama.

Misalkan G grup, dan $e_1, e_2 \in G$ merupakan unsur identitas pada G. Akan ditunjukkan $e_1=e_2$. Karena $e_1$ dan $e_2$ unsur identitas, maka untuk sebarang $a \in G$ berlaku:
$\begin{align*} &e_1a=a &\textbf{(1)} \\ &ae_2=a &\textbf{(2)} \end{align*}$

Dengan mengganti a pada persamaan (1) dan (2) secara berturut-turut dengan $e_2 $ dan $e_1$ diperoleh
$\begin{align} & e_1e_2=e_2 & \textbf{(3)} \\ & e_1e_2=e_1 & \textbf{(4)} \end{align}$

Berdasarkan persamaan (3) dan (4), diperoleh $e_1=e_2$. Jadi, terbukti bahwa unsur identitas pada grup bersifat tunggal.

Teorema 2: Invers anggota pada suatu grup bersifat tunggal.

BUKTI:  Ide yang digunakan dalam membuktikan teorema ini serupa dengan ide pada teorema sebelumnya. Tinjau sebuah anggota dari grup, misalnya x. Untuk membuktikan bahwa invers dari x bersifat tunggal, kita dapat menunjukkan bahwa jika dua objek merupakan invers dari x maka dua objek tersebut haruslah sama.

Misalkan G grup, dengan unsur identitas e. Ambil sebarang $a \in G$, dengan $b, \ c \in G$ merupakan invers dari a. Akan ditunjukkan b=c. Karena b dan c invers dari a, maka berlaku:
 $\begin{align} & ba=e & \textbf{(1)} \\ & ac=e & \textbf{(2)} \end{align}$

Perhatikan bahwa:
$\begin{align} b &= be & \text{[e unsur identitas]} \\ & = b(ac) & \text{[Berdasarkan (2)]} \\ &= (ba)c & \text{[Sifat asosiatif]} \\ &= ec & \text{[Berdasarkan (1)]} \\ &= c & \text{[e unsur identitas]} \end{align}$

Diperoleh b=c. Jadi, terbukti bahwa invers anggota pada suatu grup bersifat tunggal.

Baca juga: Memahami Ketunggalan Unsur Identitas dan Invers

Teorema 3: Jika (G,*) adalah suatu Grup maka berlaku :

i) $(a^{-1})^{-1}=a$ untuk setiap $a \in G$

ii) $(a*b)^{-1}=b^{-1}*a^{-1}$ untuk setiap $a, b \in G$

Sebelum kita buktikan, pahami dulu maksudnya. Contoh Misal kita punya himpunan bilangan bulat (Z) anggotanya { . . . , -3, -2, -1, 0 , 1, 2, 3, . . .} telah dibuktikan pada tulisan sebelumnya bilangan bulat dengan operasi penjumlahan biasa (+) membentuk grup. Sekarang, karena (Z,+) grup maka berdasarkan Teorema 1 pasti sebarang anggota a di Z berlaku $(a^{-1})^{-1}=a$. 

Contoh a=3, invers penjumlahan dari a=3 adalah $a^{-1}=-3$. Kita lihat bahwa $(a^{-1})^{-1}=3$ karena invers penjumlahan dari -3 adalah 3.

Untuk yang bagian ii), kita coba misalkan a=3 dan b=4 maka $(a+b)^{-1}=-b+(-a)=-4+(-3)=-7$. Ternyata benar, invers penjumlahan dari (3+4) adalah -7.

Catatan: $a^{-1}=-a$ karena operasi yang kita gunakan adalah operasi + biasa. Kalau operasi yang kita gunakan adalah perkalian biasa (x) maka $a^{-1}=1/a$.

Bukti Teorema 3:

i) Karena (G, *) Grup maka perhatikan:

$(a^{-1})^{-1}*a^{-1}=e$ dan pada sisi lainnya $a*a^{-1}=e$, dari sini kita simpulkan $(a^{-1})^{-1}=a$.

ii Karena (G,*) grup maka:
$1) \ (a*b)^{-1}*(a*b)=e$
$ \begin{align*} 2) \ (b^{-1}*a^{-1})*(a*b) &= b^{-1}*(a^{-1}*a)*b \\ &= b^{-1}*e*b \\ &=(b^{-1}*e)*b \\ &=b^{-1}*b \\ &=e \end{align*}$.

Jadi berdasarkan 1) dan 2) $(a*b)^{-1}=b^{-1}*a^{-1}$.

Teorema 4Misalkan G grup dan a, b, c sebarang anggota dari G. 

a. Jika ab=ac maka b=c. 
b. Jika ba=ca maka b=c. 

BUKTI: Misalkan G grup. Ambil sebarang $a,b,c \in G$. 

Bagian a. Diketahui ab=ac. Akan dibuktikan b=c. Perhatikan bahwa 
 $\begin{align} ab &= ac \\ a^{-1}(ab) &= a^{-1}(ac) \\ (a^{-1}a)b &= (a^{-1}a)c &\text{[Sifat asosiatif]} \\ eb &= ec &[a^{-1} \text{ invers dari a]} \\ b &= c &\text{[e unsur identitas]} \end{align}$  

Bagian b. Diketahui ba=ca. Akan dibuktikan b=c. Perhatikan bahwa 
$\begin{align} ba &= ca \\ (ba)a^{-1} &= (ca)a^{-1} \\ b(aa^{-1}) &= c(aa^{-1}) & \text{[Sifat asosiatif]} \\ be &= ce & [a^{-1} \text{ invers dari a]} \\ b &= c & \text{[e unsur identitas]} \end{align}$  

Catatan: Teorema terakhir ini disebut sebagai hukum pembatalan (Cancellation Law). Dalam bahasa yang lebih sederhana, jika ab=ac maka unsur a yang ada pada sebelah kiri setiap ruas dapat dicoret. Secara berturut-turut, bagian a dan b dari teorema ini disebut hukum pembatalan kiri dan kanan.

Baca juga: Memahami Hukum Pencoretan

Catatan: Pembahasan tulisan Teorema 1, 2, dan 4, saya kutip dari web

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