0 votes 0 votes Set Theory & Algebra cbt-2019 + – jatin khachane 1 asked Jan 14, 2019 jatin khachane 1 955 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply jatin khachane 1 commented Jan 14, 2019 reply Follow Share why P is not subgroup of [G,*] for [P,*] 1) forall x,y belongs to P x*y belongs to P 2) identity element of G is in P i,e 1 this is satisfying right ?? 0 votes 0 votes Satbir commented Jan 14, 2019 i edited by Satbir Jan 14, 2019 reply Follow Share for Q x = 2019k +1 = 2019*0 + 1= 1 y = 2019k +1 = 2019*1 +1 =2020 x+y = 2021 ...which we cannot write in the form of 2019k + 1. where k = ...-2,-1,0,1,2,.... =>Q is not closed under subgroup H. 0 votes 0 votes jatin khachane 1 commented Jan 14, 2019 reply Follow Share For closure property it is not that if x,y belongs to SubG then x*y should belong to G It is like , if x,y belongs to SG then x*y should belong to Subgroup 0 votes 0 votes aambazinga commented Jan 14, 2019 reply Follow Share let alone P, even G is not a group. elements are not having their inverse. 1 votes 1 votes Abhinav Gupta commented Jan 14, 2019 reply Follow Share even (G,*) is not a group. inverse does not exist for every element. this question is itself wrong. 0 votes 0 votes Shobhit Joshi commented Jan 14, 2019 reply Follow Share The elements of $P$ are not having inverse, it should not be a subgroup. For $Q$, it is not having the identity element. 0 votes 0 votes Kunal Kadian commented Jan 14, 2019 reply Follow Share Here G is not a group. Inverse doesn't exists for elements of G, but question says that G is a group. 0 votes 0 votes OneZero commented Jan 14, 2019 reply Follow Share @aambazinga exactly. 0 votes 0 votes Please log in or register to add a comment.