# Lesson 34: Like Clockwork

In the previous lesson we introduced a new diagrammatic notation for depicting what a homomorphism between two groups does.  This was all very abstract in the sense that we never had to specify which two groups were under consideration, nor the exact form of the homomorphism relating the two groups.  In this lesson we will return to something a lot more concrete, and more thoroughly explore a very common and important example of a group.  In particular, we will more carefully look at the groups made up of a finite number of integers and given a “clock addition” (more technically called “modular arithmetic”) structure, as we did in the second example of the solution in Lesson 22 and in Lesson 27.  Since we have already seen this group a couple times before, the following might be rather familiar.  However, it will be good to review it carefully since it will play a vital role in the next lesson (and many other lessons to come).

We begin by considering the set $\{0, 1, 2, 3, 4\},$ where we purposefully bold the word “set” to emphasize the fact that these five elements do not have any kind of “group structure” yet—they just form a set.  Namely, we have no idea (yet) what it means to add elements in this set.  For example, it might seem natural to say that $1+2=3,$ but we quickly see that we run into trouble if we try to say that $2+4=6.$  What kind of trouble are we running into?

Well, recall that in order to form a group, we need it to be the case that our “group multiplication,” i.e., our way of taking two elements and forming a third, needs to be such that the resulting element (i.e., the “product” of this combination) is also an element in the set.  Namely, we can’t have our group multiplication take us out of our set.  For the integers $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$ this was not an issue because this set goes infinitely far in both the positive and negative directions.  Thus, whenever we add two numbers together, their sum is also some other number which is also an integer.

However, we quickly see that if we want to consider $2+4,$ which is a perfectly good sum of two elements of our set $\{0, 1, 2, 3, 4\},$ then we cannot say that the result is $6.$  The simple reason for this is $6$ is not an element of the set $\{0, 1, 2, 3, 4\}$ and this cannot be the case if we want to try to view the set $\{0, 1, 2, 3, 4\}$ as a group.

Now, to a mathematician, the natural next question is “how do we fix this problem and change our group multiplication so that this set can be viewed as a group?”  And before answering this question it is worthwhile to first address the natural next question that might arise for a non-mathematician.  Namely, that question might be “why do we care about trying to turn the set $\{0, 1, 2, 3, 4\}$ into a group?”  This is an honest question, and the answer to it is two-fold.

First, we care about turning this set into a group because it seems unlikely that this set has no added structure at all.  In particular, it might be more obvious that the set $\{Kobe, orange, 17, Democracy\}$ has no natural mathematical structure because this set truly is a collection of random objects.  However, the set $\{0, 1, 2, 3, 4\}$ is “almost” a group because there is a natural way to add (and even multiply, but we’ll come back to this later) elements together.  The only problem is that once we get to sufficiently large numbers in this set, something starts to go wrong.  In other words, there already is a “God-given” way to add some of the numbers in this set to each other, and so it may be interesting to see how to add even the numbers 3 and 4 (for example) to each other in this set.

The second reason why we want to turn this set into a group, and by far the most important reason, is simply that it is indeed super interesting to study the consequences of successfully doing so.  Namely, the results that are obtained once we do turn this set into a group are so fascinating that one gets the sense that there is something truly fundamental and amazing going on (though we won’t see all of this amazement for a little while still), and that is enough motivation to continue on.

But now that we’re all mathematicians, we don’t really need to ask why we care about doing something, we just kind of do it because it gives us something fun to think about.  So let’s see how this works.

We begin by noting that we have no problems dealing with sums like $0+1$ or $2+2,$ because these sums don’t take us out of our set.  The first issue we encounter is trying to add 3 and 2, or trying to add 4 and 1.  However, a little thought will actually show us the way out of this predicament.

Where else do we find ourselves stuck with a finite set of numbers (perhaps twelve numbers) yet also wanting to add numbers to each other indefinitely?  Let me just watch my CLOCK (hint hint) while I let you think about it for a second…

Indeed, when we tell time, we are perfectly happy adding 3 to 11 even though we only have 12 numbers to work with.  All we do is “cycle back” through to the beginning and count up from there.  Thus, if it’s 11 o-clock and we’re meeting someone in 2 hours, then we know we’re meeting them at 1 o-clock.  If it’s 12 o-clock and we’re eating breakfast in 8 hours, then we’re eating at 8 o-clock.  The important realization here is that the number 12 is in no way special at all—if we decided to only have 5 hours in a day, then our arithmetic would simply tell us that if it’s 4 o-clock and we’re meeting someone in 2 hours, then we’re meeting them at 1 o-clock.

Therefore, we see that all we need to do is the following.  Let us take the “12 o-clock” mark on the clock and simply call it zero (so that we have 0 o-clock, 1 o-clock, 2 o-clock, 3 o-clock, …, 9 o-clock, 10 o-clock, 11 o-clock, and then 0 o-clock again, and so on).  In other words, suppose that the hands of our clock pointed to 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11 instead of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.  Now suppose that instead of having twelve hours in a day, we only had five, so that the hands of our clock pointed to 0, 1, 2, 3, and 4 and then back to 0 again.  All of the sudden, this clock is resembling our set $\{0, 1, 2, 3, 4\}$ rather closely!

Indeed, we can use this fictitious clock to define a very natural way of adding numbers together in the set $\{0, 1, 2, 3, 4\}.$  Indeed, we simply let 1 plus 2 equal 3, just as it does on the clock, but then we let 4 plus 1 equal 0, just as it does on the clock as well.  Namely, 1 hour past 4 o-clock would be 0 o-clock (just as one hour past 11 o-clock is 12 o-clock on a standard clock).  Similarly, we define 2 hours past 4 o-clock to be 1 o-clock, and 3 hours past 3 o-clock to be 1 o-clock also (the hour hand goes from 3 to 4, then from 4 to 0, then from 0 to 1, so that 3+3=1).

Using this pattern, we can define addition between any two numbers in the set $\{0, 1, 2, 3, 4\}$ in a consistent way.  Namely, we simply add the two numbers together normally and then subtract off 5 to get the actual answer.  Thus, 4 plus 3 equals 7 normally, and in our clock arithmetic 4 plus 3 equals 2.  Let me emphasize that this is exactly what we do with our twelve hour clock, only we subtract by 12 instead.  Namely, 5 plus 11 is 16 normally, but 5 hours past 11 o-clock is 4 o-clock (and 4 is 16 minus 12).

One might now ask why we had to introduce the number zero into our set.  Namely, why did we have to take the analog of the twelfth hour and rename it the zeroth hour?  We did this simply so that we can have an element that, when added to any other element, leaves the other element alone.  Namely, 0 plus anything just gives the same “anything” back (i.e., 0 plus 4 is 4, 0 plus 2 is 2, etc.).

Now, one might ask why we want to have such an element, and the answer is simply that we want to make the set $\{0, 1, 2, 3, 4\}$ a group.  In particular, we know that a group needs to have an identity element, and the identity element is one such that when it is multiplied (and remember that we always mean the “abstract group multiplication,” which in this case is addition) to anything it simply gives the same anything right back.  Thus, we want to include 0 in our “clock arithmetic.”

To summarize, we claim that the set $\{0, 1, 2, 3, 4\}$ is also a group, where the abstract group multiplication is “clock arithmetic” and where the identity element is 0.  To confirm that this kind of addition (namely, where we “cycle back through the end” of the set) turns the set $\{0, 1, 2, 3, 4\}$ into a group, we need to check that each element in the group has an inverse.  Namely, we recall that in order for a set to be a group, it must have an abstract group multiplication (which we know we have now), and identity element (which we know we have now), and it must be true that each element has an inverse—an element such that, when the original element and its inverse are “abstract group multiplied,” we obtain the identity as the result.

Recall that for the integers, the inverse of any number is simply the negative of that number, because any number plus its negative gives zero, and zero is the identity in the group of integers.  Now, however, we don’t have any negative numbers in our set, so we might be worried that our elements don’t have inverses.  We shouldn’t despair, though, because our clock arithmetic yet again gets us out of this predicament!

Namely, we see that 1 plus 4 equals 0 in our clock arithmetic (because of how we cycle back through), so that 1 is the inverse of 4 (and equivalently 4 is the inverse of 1).  Similarly, 2 plus 3 equals 0 in our clock arithmetic, so that 2 is the inverse of 3 and 3 is the inverse of 2.  Finally, 0 is the inverse of 0, since 0 plus 0 is 0.  Thus, we see that in some sense 2 is equal to “negative 3,” since 2 plus 3 equals 0.  Similarly, 4 is equal, in some sense, to “negative 1,” since 4 plus 1 is equal to 0.

As a side note, let us mention that the type of addition that we have defined on the set $\{0, 1, 2, 3, 4\}$ is often called “addition modulo 5.”  The reason for this is that once we get to 5, we cycle back through to zero.  For example, 4 plus 1 (which is normally 5) is now 0, and 3 plus 3 (which is normally 6) is now 1.  In general, the result of anything “modulo 5” simply means to take the remainder of what would be obtained if we divided by 5.  Namely, 6 equals 5 times 1 with a remainder of 1, so that 6 modulo 5 is just 1.  There is more on this in this definition of modular arithmetic.

In the exercises below we will see that we can indeed define clock arithmetic on any finite set of non-negative numbers.  Namely, there is no reason why we can’t consider the same definition of modular arithmetic on the sets $\{0, 1\},$ $\{0, 1, 2\},$ $\{0, 1, 2, 3\},$ and indeed on the set $\{0, 1, 2, 3, ..., N\}$ for any positive number $N.$  Let us therefore go on to see some of these exercises and make sure that we’re very comfortable with modular, or clock arithmetic, as it will end up being very useful in the generalization that we will make in the next lesson.

Exercises

1. Consider the set $\{0, 1, 2\}$ and define clock arithmetic on this set in the usual way.  Namely, 0 is the identity element so that anything plus 0 is the same original “anything” again, 1 plus 1 is 2 as usual, but then we “cycle back through” the end of the set.  For example, 2 plus 1 is 0.  The question is, then, what is 2 plus 2?  What about 2 plus 2 plus 2?  What is the inverse of each element in this group?
2. Consider the set $\{0, 1, 2, 3, 4, 5, 6, 7\}.$  Using addition modulo 8 on this set (meaning we simply cycle back through the end of the set, as usual), determine what 5 plus 6 is.  Also determine 7 plus 7.  Finally, determine the inverse of 3.
3. Consider the set $\{0, 1, 2, 3, ..., N\}$ for some positive number $N.$  What is the inverse of the element $N-2$?

Solutions to Exercises

On to Lesson 35

Back to Lesson 33