2 votes 2 votes Set Theory & Algebra set-theory&algebra engineering-mathematics + – worst_engineer asked Nov 25, 2015 worst_engineer 5.1k views answer comment Share Follow See 1 comment See all 1 1 comment reply reboot commented Jan 29, 2021 reply Follow Share https://gateoverflow.in/250917/groups 0 votes 0 votes Please log in or register to add a comment.
Best answer 10 votes 10 votes (aba-1)=b2 (aba-1)2=b4 (aba-1)(aba-1)=b4 ab(aa-1)ba-1=b4 ab2a-1=b4 a(aba-1)a-1=b4 a2ba-2=b4 (a2ba-2)2=b8 (a2ba-2)(a2ba-2)=b8 a2b2a-2=b8 a2(aba-1)a-2=b8 a3ba-3=b8 (a3ba-3)(a3ba-3)=b16 a3b2a-3=b16 a3(aba-1)a-3=b16 a4ba-4=b16 (a4ba-4)(a4ba-4)=b32 a4b2a-4=b32 a4(aba-1)a-4=b32 a5ba-5=b32 srestha answered Nov 25, 2015 • selected Nov 25, 2015 by worst_engineer srestha comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes b32 = (((((b2)2)2)2)2 = ((((aba-1)2)2)2)2 = ............ //do recursively //(aba-1)2 = aba-1aba-1 = ab2a-1 = a(aba-1)a-1= a2ba-2 = a5ba-5 Digvijay Pandey answered Nov 25, 2015 • edited Nov 25, 2015 by Digvijay Pandey Digvijay Pandey comment Share Follow See all 4 Comments See all 4 4 Comments reply worst_engineer commented Nov 25, 2015 reply Follow Share Hi , but , as a and b are in group so , aa-1 will give identify element , right ? 0 votes 0 votes Digvijay Pandey commented Nov 25, 2015 reply Follow Share yes.. & I*A = A*I = A 0 votes 0 votes worst_engineer commented Nov 25, 2015 reply Follow Share but , I can not understand afterwards. like how (aba-1)2 is evaluated ? 0 votes 0 votes Digvijay Pandey commented Nov 25, 2015 reply Follow Share (aba-1)2 = aba-1aba-1 = ab2a-1 = a(aba-1)a-1= a2ba-2 2 votes 2 votes Please log in or register to add a comment.