Build abelian group containing a set $K$ under an associative, commutative operation $*$ with an identity but the inverses are not always in $K$.

Question

We are given a set K of elements and an operation * . For every element in the set, there exists an inverse element (not necessarily in the set). There are three (additional) rules:

1) K contains e (neutral element) such that for every x: x * e = e * x = x

2) K is associative

3) K is commutative

The task is to build an abelian group G such that G contains K OR to prove that it is impossible.

The task seems to be intuitively simple, but the concrete proof seems to be unobvious.

Answer 1

It's not necessarily possible, since you have no guarantee of the existence of inverses. Specifically, if $K$ is the set of integers and * is multiplication, then it satisfies all three rules. Alternatively, if $K = \{0,1,2,3,\ldots\}$ and * is addition, you also have all three rules satisfied.

What you have with those three rules is called a (commutative) monoid. Remove condition 1) and you have a semigroup.

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Edit: After adding the rule that each element has an inverse, your (now four) rules are exactly the rules of an abelian group.