show that if a subset of $\big\{{1,2,...,21}\big\}$ contains an even number or contains the number $11$, then it cannot form a group under multiplication modulo $22$.
first of all it's clear that the multiplication of the two numbers $2$ and $11$ is a multiple of $22$, implies $2·11≡0 (mod\: 22)$, since the element $0$ is not a member of the given set so the group is not closed under the binary operation. but how is it possible to find all the pair of numbers which their multiplication modulo $22$ has a reminder which doesn't exist in the given set (assume I could not find $2$ and $11$).
also $2$ and $11$ does not have any inverse, means there is no element which holds in the equation $2·x≡1 (mod\: 22)$ and the same about $11$, but how we can be sure about that?
It is enough to show that not all elements are invertible, that means the set cannot be a group with respect to this operation. How can we be sure that $2$ of $11$ are not invertible? Well, let's assume $11$ is invertible, i.e there is $x\in\{0,1,...,21\}$ such that $11x\equiv 1$(mod $22$). By definition this means that $22|(11x-1)$ and hence there is $y\in\mathbb{Z}$ such that $11x-1=22y$, or equivalenly $11x-22y=1$. But note that this equation implies that any integer which divides both $11$ and $22$ must divide $1$ and hence we get $\gcd(11,22)=1$ which is of course a contradiction. So $11$ is not invertible mod $22$.
You can do the same thing for $2$, and in general $a\in\{0,1,...,n-1\}$ is invertible mod $n$ if and only if $\gcd(a,n)=1$.