So, you have a monoid, that is, a set with an associative binary operation that has an identity element. But not all elements have inverses, so it is not a group. Assuming you really want a group, what can you do?
It turns out there is a very simple answer: just throw away all the elements that don’t have inverses! That is, given a monoid , form the subset of all the elements of that do have an inverse with respect to the binary operation. I claim that is in fact a group (under the same binary operation). Let’s prove it. We have to check that satisfies all the laws of a group.
 By assumption, there is some which is the identity for the binary operation. So in particular , which means that is its own inverse; so .
 We know all the elements in have an inverse in , but we still need to make sure those inverses actually end up in ! If then by definition has an inverse, , which means that . Note that these equations are completely symmetric: not only is the inverse for , we can also say that is the inverse for . So must be in too.
 If the binary operation is associative for then it will definitely be assocative for too (since associativity holds for all elements of , of which is a subset).

There is one more crucial thing we have to check: it is not a priori clear that is even closed under the binary operation—might there be some such that is in but not in ?. But in fact if and both have inverses, then must have an inverse as well, namely, , since
.
(The fact that is sometimes called the “socksshoes property”: you put on your socks first (), and then your shoes (); the inverse operation is to first take off your shoes (), then your socks ().)
So if and are in then must be as well.
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