1 votes 1 votes Set Theory & Algebra discrete-mathematics group-theory zeal zeal2019 + – Prince Sindhiya asked Dec 21, 2018 • retagged Nov 29, 2023 by Hira Thakur Prince Sindhiya 511 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Satbir commented Dec 21, 2018 reply Follow Share 3? 0 votes 0 votes kumar.dilip commented Dec 21, 2018 reply Follow Share I think only one (i). 0 votes 0 votes Satbir commented Dec 21, 2018 reply Follow Share 4th option is a cyclic group ? @kumar.dilip 0 votes 0 votes OneZero commented Dec 21, 2018 reply Follow Share 1 and 4 0 votes 0 votes kumar.dilip commented Dec 21, 2018 reply Follow Share Satbir I'm sure about 1 but not 4. ( I need to revise this topic). 0 votes 0 votes Prince Sindhiya commented Dec 21, 2018 reply Follow Share how you got 4th 0 votes 0 votes Satbir commented Dec 21, 2018 reply Follow Share Option 1 is correct coz it follows closure, identity is 1, invere of every element exist and multiplication follows associativity. Option 2 is also correct coz it follows closure, identity is 0, each element has inverse and associativity is applicable on addition. Option 3 does not follows associativity. Option 4 is a cyclic group i think....every cyclic group is abelian group and thus it is also a group. 0 votes 0 votes Prince Sindhiya commented Dec 21, 2018 reply Follow Share 2) closer not satisfy 1+1=2 which is not in set 0 votes 0 votes Please log in or register to add a comment.