43 votes 43 votes The set $\{1,2,3,5,7,8,9\}$ under multiplication modulo $10$ is not a group. Given below are four possible reasons. Which one of them is false? It is not closed $2$ does not have an inverse $3$ does not have an inverse $8$ does not have an inverse Set Theory & Algebra gatecse-2006 set-theory&algebra group-theory normal + – Rucha Shelke asked Sep 16, 2014 Rucha Shelke 9.7k views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Sankalp Singh 1 commented Nov 18, 2020 reply Follow Share In order to find the inverse, we need to find the identity element first. Here what will be the identity or even the identity exist? 0 votes 0 votes HitechGa commented Jul 14, 2021 reply Follow Share While solving this question for the first time, I had a weird feeling about the meaning and options of the questions. The definition of group which I had in the back of my mind, rather the algorithm which I use to check whether a set and a binary operation is a group or not is as follows: A monoid with identity element $e$ is a group, iff for each element $a$ in the monoid, there exists as element $x$ such that $a*x=x*a=e$. So $x$ is the inverse of $a$ i.e. $x=a^{-1}$. So intuitively, in the process of checking, if a structure failed to be an algebraic structure [closure property not followed], then it is not a group obviously. And this step-by-step checking of structures made me feel that $(A)$ is the only true option for the given structure to be not a group, Rest all are false. [Hence I thought correct options are BCD] But the alternate definition of groups (in texts which talks about groups directly without talking about algebraic structure, semi group, monoids) says: A set $S$ with a binary operation $*$ defined on its elements is a group iff (1) $S$ is closed under $*$ and (2) $*$ is associative and (3) there exists an identity element and (4) There exists inverse for each element. From this alternative definition, since the requirements for being a group is given in AND form, then for NOT being a group the requirements shall be : $F = \lnot(1) \lor \lnot(2) \lor \lnot(3) \lor \lnot(4)$ Although if anyone of $(1)$ or $(2)$ or $(3)$ or $(4)$ is false then $F$ shall be true, but strictly speaking there is no precedence or order among the terms of $F$. They can just be rearranged and hence options $B$ and $D$ are also true since make $\lnot(4)$ true. 5 votes 5 votes Akash 15 commented Dec 27, 2023 reply Follow Share Given set $S=\{1,2,3,5,7,8,9\}$ $(\{1,2,3,5,7,8,9\},\bigotimes_{10})$ is not a group. Option by option: It is not closed $(2\times 3) mod \;10=6 \notin S$ $(2\times 5) mod \;10=0 \notin S$ $(2\times 2) mod \;10=4 \notin S$ So we can observe $\{0,4,6\}$ these are not in $S$. This group is not closed as they not belong to our base set. $2$ doesn’t have an inverse $(2\times 1) mod \;10=2 \in S$ $(2\times 2) mod \;10=4 \notin S$ $(2\times 3) mod \;10=6 \notin S$ $(2\times 5) mod \;10=0 \notin S$ $(2\times 7) mod \;10=4 \notin S$ $(2\times 8) mod \;10=6 \notin S$ $(2\times 9) mod\;10=8 \in S$ So, here is no identity element. Thus the inverse of $2$ not even exist. $3$ doesn't have an inverse No. $3$ has an inverse, which is $7$ $(3\times 7)mod\;10=1;\;\;(7\times 3)mod\;10=1\;\;\therefore7=3^{-1}$ $8$ doesn’t have an inverse $(8\times 1) mod \;10=8 \in S$ $(8\times 2) mod \;10=6 \notin S$ $(8\times 3) mod \;10=4 \notin S$ $(8\times 5) mod \;10=0 \notin S$ $(8\times 7) mod \;10=6 \notin S$ $(8\times 8) mod \;10=4 \notin S$ $(8\times 9) mod\;10=2 \in S$ So, here is no identity element. Thus the inverse of $8$ not even exist. $\color{DarkGreen} Only\;C\;is\;false$ 1 votes 1 votes Please log in or register to add a comment.
Best answer 33 votes 33 votes Answer: C $3$ has an inverse, which is $7.$ $3*7 \mod 10 = 1.$ Rajarshi Sarkar answered May 11, 2015 selected Jun 2, 2015 by Rajarshi Sarkar Rajarshi Sarkar comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments raja11sep commented Oct 18, 2022 reply Follow Share also 8*8 mod 10 = 4 not in list 0 votes 0 votes Pranavpurkar commented Oct 21, 2022 reply Follow Share LoL! 0 votes 0 votes Akash 15 commented Dec 27, 2023 i edited by Akash 15 Dec 27, 2023 reply Follow Share also 7*8 mod 10 = 6 not in list Follow this trend :) If you want option analysis then you can look this: https://gateoverflow.in/882/gate-cse-2006-question-3?show=417093#c417093 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes Ans C. Vikrant Singh answered Dec 31, 2014 Vikrant Singh comment Share Follow See all 2 Comments See all 2 2 Comments reply anshu commented Feb 4, 2015 reply Follow Share How?? 0 votes 0 votes Sandeep Suri commented Jan 11, 2017 reply Follow Share In the question it is asked which is false. and given that it's not a group. As we can see 3 have an inverse which is 7 therefore this statement is false 1 votes 1 votes Please log in or register to add a comment.
3 votes 3 votes Hi , please correct me if I am wrong. The set is {1,2,3,5,7,8,9} and we need to do multiplication modulo 10. So, 2 (multiplication modulo 10 ) 2 = 4 , which is not in the set. That means , it is not closed. Please correct me , if I am wrong . worst_engineer answered Jul 18, 2015 worst_engineer comment Share Follow See all 3 Comments See all 3 3 Comments reply vishal8492 commented Jul 20, 2015 reply Follow Share You're right 0,4,6 are missing ; so for every n mod 10 = {0,4,6} this set is definitely not closed. But trick is it is not closed is not false reason ; it's True. 8 votes 8 votes worst_engineer commented Jul 20, 2015 reply Follow Share Ohh yeah.. sorry didn't see the question properly. my bad :( . Yes , it is true. thanks 0 votes 0 votes Sandeep Suri commented Jan 9, 2018 reply Follow Share Question is asking which of the statement is not FALSE.. Yes it is not closed is a True statement. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes . akshay_123 answered Sep 2, 2023 akshay_123 comment Share Follow See all 0 reply Please log in or register to add a comment.