GATE CSE 2006 | Question: 3

Question

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?

• A. It is not closed
• B. $2$ does not have an inverse
• C. $3$ does not have an inverse
• D. $8$ does not have an inverse

Comments

8*7=56%10=6..not in the set ...so it is not closed ...why it cant be the anssssss

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@asu Bhai question is bit tricky asked

they are asking which one is false.

Here 3 having inverse which is 7 .bcz 3*7=21mod10=1 //1 is identity element

but they are telling 3 does not have an inverse. So false.So C is Ans.

(8 does not have an inverse.This statement is true..but they asked for false. so it cant be ans)

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I think there is bit of confusion in here, this is not group because it is not closed at first so why we to talk about Inverse, it doesn’t matter and is not at all reason for this to be not a group, the reason is it is not closed. So I think it is not dependent on option B,C,D and talking about inverse is meaningless in here.

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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?

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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.

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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:

1. 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. $2$ 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.

100. $3$ doesn't have an inverse

No. $3$ has an inverse, which is $7$ 

$(3\times 7)mod\;10=1;\;\;(7\times 3)mod\;10=1\;\;\therefore7=3^{-1}$

500. $8$ 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.

$\color{DarkGreen} Only\;C\;is\;false$ 

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5*8 = 40 Mod 10 = 0, which is not in the group. It is not closed.

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question is mindblowing only because of the framing that the options are true and asked for false stmt .But in hurry none cares

Answer 1

Answer: C

$3$ has an inverse, which is $7.$
$3*7 \mod 10 = 1.$

Comments

also 8*8 mod 10 = 4 not in list

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LoL!

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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

Answer 2

Ans C.

Comments

How??

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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

Answer 3

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 .

Comments

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.

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Ohh yeah.. sorry didn't see the question properly. my bad :( . Yes , it is true. thanks

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Question is asking which of the statement is not FALSE.. Yes it is not closed is a True statement.

Answer 4

$\begin{array}{|c|ccccccc|} \hline
⊗_{10}&1&2&3&5&7&8&9\\ \hline
1&1&2&3&5&7&8&9\\ \hline2&2&4&6&0&4&6&8\\ \hline3&3&6&9&5&1&4&7&\\ \hline5&5&0&5&5&5&0&5&\\ \hline7&7&4&1&5&9&6&3&\\ \hline8&8&6&4&0&6&4&2&\\\hline9&9&8&7&5&3&2&1\\ \hline
\end{array}$

from the above composition table, it is clear that $1$ is the identity element.

1. it is not closed: true, because elements like $0,4,6\notin$ the set $\left \{ 1,2,3,5,7,8,9 \right \}$
2. $2$ doesn't have an inverse: true
3. $3$ does not have an inverse: false, inverse (3)=7
4. $8$ does not have an inverse: true.

So Option $(C)$ is the correct.

Comments

nice one

Answer 5

.

Answer 6

Why option A is correct ?? we know possible remainder when divide by n is 0,1,2,3,4 .... n-1 which is not in Set S so it is not  covering all remainders properly.