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.

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 :

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.

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.