 Well, let's take a look at another multiplication table for the group with four elements. Again, part of being mathematically sophisticated is to not stop when you found one answer, but to look for all possible answers. So just a reminder, we found a multiplication table for a group with four elements, namely the identity E and the elements A, B, and C, and we found that by boarding one particular bus, the A times A equal E bus. So we put that there, and once we boarded this bus, we found a bunch of other products that were necessary, but at some point the bus ride ended and we had to make another choice. We had to get on a different bus, and so we boarded at that point the B times B equal to E bus, and we were able to board and ride that bus to the end of the line. Now, you'll notice that these are color-coded here. The black are absolutely necessary because we have a group. These are just the products with the identity. The green all come from this initial assumption, A times A equals E, and then the blue come from this initial assumption, B times B equal to E, and it's helpful to do that because it means that when we change our assumption, we can eliminate all of the products of one particular color. So if I wanted to change this assumption here, I'd eliminate all of these. If I wanted to change this assumption here, I could eliminate all of the green, and I can also just reset the entire table. So let's use that color coding. If you're not using colored pencils, you can use some other form that will make it easy to identify what follows from which particular bus you board. So let's get on a different bus. Alright, so first of all, the identity and the products are going to remain the same, but this time let's board the A times A equals B bus. So we originally boarded the A times A equals E bus. Let's board the A times A equals B bus, and there's my product. And now I want to find what the product A times B is going to be. So let's look at our possibilities. A times B could be the identity. Nothing wrong with that. A times B equals A can't happen because B is not the identity. A times B equals B again can't happen because A is not the identity. A times B equals C, no obvious problems with that. So this is actually a pretty short bus ride. We made this initial assumption A times A equals B, and we only got as far as this next box, and now we have one of two buses that we could get onto. So let's pick one, and this time let's board the A times B equals C bus. Now remember G is a group, which means that A times something has to give us the identity, not this, not this, not this. Our only choice is A times C has to give us the identity. So this last product here is going to be E, and also the identity commutes. Once I know that A times C is equal to E, I also know that C times A has to be E as well. So I also have this value filled in, and notice that these three values are all connected to each other. All right, well what's next? We want to see if we can get to the B times A product. So again, looking at our possibilities can't be the identity because A already has its identity, and it's not going to be B. B times A can't be A, can't be B. The only possibility left, it has to be C. B times B, well B times B could be the identity, no obvious problems with that. B times B equals A, no obvious problems with that either. B times B can't be B, can't be C. Now you might wonder why it can't be equal to C. Well remember that we already have B times A equals C. If we also have B times B equal to C, that's going to require that A and B be the same element, and we know that's not true. So it can't be B, can't be C, and it seems like this bus ride only got us to this point, and now we have to board another bus. So let's board the B times B equals A bus, and see where that takes us. All right, so as before B needs an inverse. So I know that B times something has to give us the identity, and that means that B times C has to give us the identity. Except there's a problem because B times C, if B times C is E, A times C is also E, and so B times C and A times C are the same thing, and right cancellation tells me B and A are equal, but they can't be equal. They're supposed to be different elements. I'm supposed to have four elements in the group. So this can't be allowed to happen. B times C can't be equal to E, and our bus crashes. Maybe we don't want to get on this particular bus then. So we got on the bus B times B equals A, it crashed, and so let's not get on that bus. And it said take the other bus B times B equals E. And so finally B times C, we'll go through our possibilities, can't be E because we've already found the inverse of B. Maybe it's A. Can't be B, C is not the identity. Can't be C, B is not the identity. So our only possibility B times C has to be the element A. C times B, well again we know it can't be the identity because we already have the inverse for C, could be A, can't be B, C is not the identity, can't be C, B is not the identity. So our only possibility C times B has to be A. And finally C times C can't be the identity, could be A, except this runs us into the same problem because I know that B times C is A. If I have C times C is A, again right cancellation is going to tell us that B and C are the same and I don't want that. So it can't be A, could be B, and it can't be C because again C is not the identity elements. Our only choice that last product has to be B. And this completes a different multiplication table for a group with four elements.