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

### 5 Comments

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)

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.