 Another important group emerges when we study the set of integers mod n. Suppose we consider what happens when the integers are divided by a specific number n. n is called the modulus, and the possible remainders when we do this division are the numbers 0, 1, 2, 3, and so on, all the way up to n minus 1. If two numbers have the same remainder when divided by n, we write a equivalent to b, or sometimes we read this as a is congruent to b. This assumes that everybody knows what n is. If not, if we have reason to have to specify n, we should indicate that in some fashion. So we might write it this way, a is congruent to b, and somehow indicate that there's an n there, or we might write a is congruent to b taken mod n, or we might even write a mod n is congruent to b mod n. But the most common way of writing this is to write a congruent to b mod n. The most important thing to notice about this notation is that the mod n applies to both sides of the congruent symbol. It does not apply just to b, it applies to the a as well. Now, this set of remainders is designated zn, and if I want to form this into a group, I have to choose an operation. Whether I want to use plus or whether I want to use times, I could use either, and soon enough we'll use both, but we'll see plus has certain advantages. Before we proceed, we need to introduce an important lemma. Suppose I'm working mod n, and I'm told that a is congruent to b. Then that a minus b is going to be a multiple of n. To prove this, it's useful to remember if I find m divided by n equal to q with remainder r, then I know that m is equal to qn plus r. Now, if a is congruent to b, then they have to have the same remainder when divided by n. So I know that a divided by n is some quotient with some remainder, and so a is equal to q1n plus whatever that remainder is. I also know that b divided by n is some quotient with the same remainder, and so that tells me that b is q2n plus the same remainder. And I want to prove something about a minus b, so I'll subtract the two and do a little bit of algebra, and I find that a minus b is going to be n times the difference of two numbers. And so that tells me that a minus b is going to be a multiple of the modulus n. So here's an important property you should be able to prove. Suppose I have a, b, c, and d, all integers with a congruent to b and c congruent to d all taken mod n. Then if I add a and c, it's going to be congruent to what I get by adding b and d. And likewise, if I multiply a by c, I should get something congruent to the product of b and d. So let's see if we can form the Cayley table for the group of integers mod 9 using the operation of multiplication. Now, first of all, a group must include the identity, and since the operation is multiplication, our identity is going to be 1. And because every element of a group must have its inverse, then I can only include an element a if I can find its inverse b. In other words, a, b has to be congruent to 1. Well, our lemma says that if two things are congruent, their difference is going to be a multiple of the modulus. So in this case, our modulus is 9. So that means a, b minus 1 has to be a multiple of 9. So let's see. Well, we find that 2 has inverse 5 because 2 times 5 minus 1 is 9, which is self-evidently a multiple of 9. This also means that 5 is a group element. 3, on the other hand, does not have an inverse because you can't find a value for b for which 3b minus 1 is going to be a multiple of 9. How about 4? Since we know inverses are unique, we know the inverse of 4 is not 2 or 5, and since 3 is not even in the group, it's definitely not going to be 3. So that means we have to look at the other numbers, and with a little bit of trial and error, we find that the inverse of 4 has to be 7 because, again, 4 times 7 minus 1 is 27, which is, again, a multiple of 9. 6, again, is not something that has an inverse because we cannot solve 6b minus 1 as a multiple of 9. We already know 7 is in the group because it's the inverse of 4. 8 can't have inverse 2, 4, 5, or 7. What about 8? Well, we find that 8 times 8 minus 1 is 63, and that is, in fact, a multiple of 9. And so 8 is its own inverse. And finally, 0, which is a possible remainder, can't have an inverse because 0 times anything minus 1 will never be a multiple of 9. And so let's go form our Cayley multiplication table. And we know that 1 is the identity, so we can fill out that row and column. We know that 2 and 5 are inverses, so we know 2 times 5 and 5 times 2 are both equal to 1. Likewise, 4 and 7 are inverses, and 8 is its own inverse. So the key here is to remember that a is congruent to b whenever a and b have the same remainders, so that allows us to find the remaining products. So 2 times 2 is 4. 2 times 4 is 8. 2 times 7 is 14, but 14 is not a group element, but it is congruent to 5. So 2 times 7 is 5. 2 times 8 is 16, which is congruent to 7, so 2 times 8 is equal to 7. Now, remember, the times here is the multiplication of integers, and integer multiplication is commutative, so at the same time that we have 2 times 2, 2 times 4, 2 times 7, 2 times 8, we also have 4 times 2, 7 times 2, and 8 times 2, so we can fill in those as well. Likewise, 4 times 4 is 16, which we can reduce. 16 is congruent to 7. 4 times 5 is 20, which is congruent to 2. 4 times 8 is 32, which is congruent to 5. And commutativity allows us to find three more products. 5 times 5 is 25. That's going to be congruent to 7. Again, because 25 minus 7, 18, is a multiple of 9. Likewise, 5 times 7 is 35, congruent to 8. 5 times 8 is 40, congruent to 4. And that allows us to fill in some more of our Cayley table. And finally, 7 times 7, 49, which is congruent to 4. 7 times 8, 56, which is congruent to 2. And so here is the group of integers mod 9 where we use the operation of multiplication. And one of the disquieting things about this is that while there are other elements of the integers mod 9, we can't include all of them in the group, and that's because we're doing multiplication and not addition. So, preferentially, when we talk about the group of integers mod n, we want to focus on using the operation of addition, unless otherwise specified.