 In this video, we're going to prove the so-called cancellation laws of a group, and they're stated in the following way. Let G be a group, and take elements A, B, C. These are just arbitrary elements in the group, but they have the condition that B times A is equal to C times A. Notice that A is the same thing on the back here. So, if A, B equals C, A, this actually implies that B equals C. The only way that B, A can equal C, A is that B actually equals C in the first place. This property right here we call right cancellation. You can cancel A on the right side of the equation. And similarly, if A, B were to equal A, C, we could actually then infer from that that B equals C. This is an example of what we call left cancellation, that we can cancel out A on the left-hand side. These are very useful properties in algebra. This is actually a property of being a group. So, what we're going to do is we're actually going to prove left cancellation. I'm going to prove left cancellation and argue that right cancellation is similar. So, this is actually a conditional statement right. So, if A, B equals A, C, that implies B equals C for any elements of the group. So, we're going to start off by assuming that A, B equals A, C. So, this is our starting point. Now, using properties of groups, we're going to prove that B actually equals C. And so, the way we're going to do that is we're going to multiply both sides of the equation by A inverse. Because A belongs to the group, it has an inverse. So, we're going to use the inverse axiom to have existence of that group element. Now, because the group is, well, a group, we can re-associate the parentheses. The associative property says that we can redo the parentheses so that I'm actually going to instead look at A inverse A times B and A inverse A times C. Since these are inverses, they multiply A inverse A is going to give us an E, which we see right here. And because E is the identity, so we have the identity element right here, E, B is just B and E, C is just C. So, we were able to cancel A on the left-hand side of these equations. So, if A, B equals A, C, then B actually was equal to C. We have left cancellation. Right cancellation is proven very similar to this. And so, I want you to realize that here that, in fact, the above proof used all three axioms of group theory. So, associativity, inverse as identity as I illustrated here on the screen. In some regard, any object with cancellation must be a group-like object. And in fact, if you have a set with a binary operation that satisfies the cancellation laws, we actually call this a quasi-group. The idea is that it's kind of like a group. If you have cancellation, you're very close to being a group. But basically, quasi-groups don't require associativity, which is sort of like a fatal mistake. Associativity is the best axiom out there. Things get really funky when you don't have associativity. But hey, once you have the cancellation laws, I want to mention that you can get stuff from that, right? We can actually solve equations uniquely. I'll say we can solve linear equations uniquely inside of a group. If we have a group G, take elements AB, these are elements in the group. Then the equation ax equals B, which here x is a variable, we don't know what it is. So A and B are fixed, but x is a variable. The linear equation ax equals B and xA equals B have unique solutions in the group. I have to write both of them because the group isn't necessarily commutative. But I'm going to give you an argument that ax equals B has a solution, a unique solution. And by similar reason, xA equals B will have a solution as well. And you're going to see that this is going to basically follow from the cancellation laws. So if you have an equation, let's first prove that it has a solution in the first place, right? What's the solution to ax equals B? Well, if you're looking at this ax equals B, you'd be tempted just to multiply both sides of the equations by a inverse, kind of like we did with the cancellation laws, and you would get that x equals a inverse B. Because the operation is non-commutative, you have to make sure you put the inverse on the left side. To cancel a on the left, your inverse needs to be on the left of B, not on the right. That would not give you the right element. So my candidate for the solution is actually x equals a inverse B. And so what we can do is we can then plug that in for x, and you'll see it right here. If you replace x with a inverse B, you'll get a times a inverse B. Reassociate, you get a times a inverse times B. A times a inverse is the identity, and the identity times B gives you just a B, right? We used all the axioms again. Existence of inverses was used, associativity was used, the identities were used. We used all the axioms of groups in that statement. This just proves the existence, the existence of solution. So this just tells us that there is a solution, but it doesn't tell us that there's a unique solution, right? The uniqueness usually has to be supplied as a second argument. So how do we get uniqueness of a solution? Why can't there be a second solution? Well, if there's a second solution, this is where cancellation is going to come into play. Say there's two solutions, both x and y. Well, if x is a solution, that means ax equals B. If y is a solution, that means ay equals B. But if they both equal B, that actually means that ax is equal to ay. By left cancellation, you're going to get that x equals y. So uniqueness is a consequence. The uniqueness of the solution is a consequence of the cancellation that we saw just above. And so because of the group axioms, which cancellation is a consequence, we can prove that linear equations have unique solutions. Ax equals B and xa equals B. xa equals B, of course, is handled similarly.