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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?

1. It is not closed
2. $2$ does not have an inverse
3. $3$ does not have an inverse
4. $8$ does not have an inverse

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

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.

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

The given relation is not closed also, 2*8 mod 10 = 6 which doesn't belong to the set. Can it be a reason for it not being a group?
yes , it is one of the reasons.
also 5*2 mod 10 =0 not in list
also 8*5 mod 10 =0 not in list.
Ans C.

How??
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
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 .

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.

Ohh yeah.. sorry didn't see the question properly. my bad :( . Yes , it is true. thanks
Question is asking which of the statement is not FALSE.. Yes it is not closed is a True statement.