The question is Let G and G\'\' be groups (with a,b,c in G and a\',b\',c\' in G\
ID: 2938800 • Letter: T
Question
The question isLet G and G'' be groups (with a,b,c in G and a',b',c' in G'') where* denotes the operation of both groups. Define * on G X G''(G cross G'') so that (a, a') * (b, b') = (a*b, a'*b'). Briefly explain why GxG'' with this * is a group.
I know the three axioms for a group (associativity, identity andinverse), and I sort of understand the concept of a Cartesianproduct. But I am just completely stuck. Can anyonegive me a hint or a line of thinking that would help me start insolving this problem?
Explanation / Answer
The question is Let G and G'' be groups (with a,b,c in G and a',b',c' in G'') where* denotes the operation of both groups. Define * on G X G''(G cross G'') so that (a, a') * (b, b') = (a*b, a'*b'). Briefly explain why GxG'' with this * is a group. G=[A,B,C].....................G'=[A',B',C'] IF 3 ELEMENTS ARE IN ONE GROUP .THEN ONE OF THEM IS I...... SAY A=I WITHOUT ANY LOSS OF GENERALITY. HENCE A*B=B...A*C=C.. THEN INVERSE OF B IS C AND SO B*C=C*B=A SO WE HAVE B*B=C AND C*C=B SO THE COMPOSITION TABLE FOR G IS A B C A A B C B B C A C C A B SIMILAR TABLE APPLIES FOR G' A' B' C' A' A' B' C' B' B' C' A' C' C' A' B' GXG'=[(A,A'),(A,B'),(A,C'),(B,A'),(B,B'),(B,C'),(C,A'),(C,B'),(C,C')] HERE THERE IS NO SIMPLE WAY TO GO BY WE HAVE TO MAKE THE COMPOSITION TABLE AND PROVE THE 3 RULES YOUMENTIONED. USING THE ABOVE COMPOSITION TABLES FOR G AND G' HERE WE FIND THAT A,A' IS THE IDENTITY ELEMENT SINCE A AND A' AREIDENTITY ELEMENTS AND (A,A')X(Z,Z')=[(A*Z),(A'*Z')]=[Z,Z'] IN GENERAL ....Z IS COMMONREPRESENTATION FOR B OR C. MAKE A TABLE AS GIVEN BELOW AND YOU WILL FIND ALL YOUR REQUIREMENTSMET IT IS PURE LEG WORK IF STILL IN DIFFICULTY PLEASE COME BACK A,A' A,B' A,C' B,A' B,B' B,C' C,A' C,B' C,C' A,A' A,A' A,B' A,C' B,A' B,B' B,C' C,A' C,B' C,C' A,B' A,B' A,C' A,C' B,A' B,A' B,B' B,B' B,C' B,C' C,A' C,A' C,B' C,B' C,C' C,C'
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