 Given an algebraic structure with an identity, one question we could ask is could G star have more than one identity element? To answer this question, remember that definitions are the whole of mathematics, and an identity element is one that if it operates on any element G, it's going to give us G itself. Now a useful proof strategy in abstract algebra, and many other areas of mathematics, suppose there are two. So suppose E and E prime are both identity elements. Definitions are the whole of mathematics, all else is commentary, let's pull in our definition of identity. Now since E is the identity element, we know that E star E prime has to give us E prime. But wait, if E prime is the identity, we know that E star E prime has to give us E. But our products are both E star E prime, and so that means the results E prime and E have to be equal, and so this gives us the following theorem, let G star be an algebraic structure, if it exists, the identity is unique. So remember that a semi-group with an identity is a monoid, and so we might ask the question, since a monoid is a semi-group and semi-groups don't require the Latin square property, what happens if we include it? Well let's find out. Let our group elements be E, X, Y, Z, and so on, where E is our identity element. We'll also assume that G has a finite number of elements. Suppose A is some element of G, and we'll use the notation A star G to be the set A star X, A star Y, and so on. In other words, we're going to left multiply every element of G by A, and we'll define G star A to be where every element of G is right multiplied by A. Now if G star has the Latin square property, then the elements of A star G are distinct and must include every element of G. And so at least one of them must be the identity, giving us A star P equal the identity for some element P in G, and similarly there must be some element Q in G for which Q star A gives us the identity element. And it's useful to keep in mind that we want to be able to solve equations, we can use these to solve equations. So suppose we've already found these, and so we know that A star X is the identity, and also Y star A is the identity. And suppose we have an equation A star something equals B. So as long as we do the same thing to both sides, we still have an equation, so let's left multiply by Y. Now it's important to understand here that A star Z is supposed to be a single thing, so when we left multiply by Y, it's really Y star the quantity A star Z. And because we're left multiplying, we also have to left multiply by Y on the right hand side and get Y star B. But for now we'll assume that we're working with a semi-group and we have associativity. Remember if we don't have associativity things get a lot more complicated, and here's one of the reasons why. Because we have associativity, we can associate Y and A and then leave Z out. But we know what Y star A is, and since E is the identity, we know what E star Z is. And so we've solved for Z, we've expressed it in terms of Y, the thing that we multiplied on the left to get the identity, and B, the other term in our equation. And so we can solve any simple equation using elements of G as long as every element A has an X and Y, where A star X is the identity and Y star A is the identity. Unfortunately, this doesn't always happen. But if it does, we can be happy about it and also define some new terms. Let G star be an algebraic structure with identity E and let A be some element of G if they exist. The left inverse Y satisfies Y star A equals the identity, while the right inverse X satisfies A star X equals the identity. Now, the hardest thing to remember about the left and the right inverses is that the left inverse only works if it's on the left side. And that should be obvious, but you do have to remember which side is the left, and so the easiest way to remember that is to hold out your hands. Your left hand is the one that forms the L. Now, sometimes the same element could be used for both a left and a right inverse. And if that happens, we say that it is an inverse, no left, no right, and designate the inverse of A to be A to power minus 1. Now, remember, there are only so many symbols, and so while this appears to be an exponent, you don't want to think about this as 1 over A, it's the inverse of A. And let's tie this back in with what we did earlier, where we found that if a monite has a finite number of elements and Latin square property, then every element has both a left inverse and a right inverse. And remember we proved that because when we multiplied everything on the left by A, something had to be the inverse, and likewise if we multiplied everything on the right by A, something had to give us the identity. But here's an important thing to keep in mind, read the fine print. While every element has a left inverse and a right inverse, we've made no claim and, importantly, provided no proof that they are the same. But let's think about that. A monite is a semi-group, and a semi-group has another important property, associativity. So what if we incorporate associativity? So suppose G star is a monite with a finite number of elements at the Latin square property. We already showed that A in G has both a right inverse X and a left inverse Y. So we know that A star X is the identity, and also that Y star A is the identity. Now, since Y is the left inverse, we'll left multiply by Y. And the important thing here is that A star X is already there. So when we star it with Y, we have to treat A star X as a single thing and closing it in parentheses. And since G star is a monite, it's also a semi-group, and so we have associativity. And so we can re-associate. Now, since Y is a left inverse, we know that Y star A is the identity. And since E is the identity, we know that E star X is just X. And that tells us X is equal to Y. And so this proves the following theorem. Suppose G star is a monite with a finite number of elements at the Latin square property, then any element of G has an inverse A inverse. No left, no right, but this inverse can be used on either side to get the identity.