Cyclic Group Intro (Math 0)
Say we had a clock. We decide that we can “add” two numbers on the clock starting at the first number, then moving by the number of steps given by the second number. For example the sum of \(5\) and \(6\) is given by starting at \(5\), and then moving by \(6\) spaces, and ending up at \(11\). But the sum of \(7\) and \(9\) is given by starting at \(7\) and moving \(9\) spaces to end up at \(4\). This number can be calculated by just adding the numbers normally, then if you end up at more than \(12\), subtract \(12\). For example, \(7+9 = 16\), and \(16- 12 = 4\).
Notice that if you add \(12\) to anything, you just go once around the clock, and so ending up not doing anything. You can calculate it with \(4\) as \(4 + 12 = 16\), and \(16 - 12 = 4\). Hence \(12\) is the identity element for our clock. Hmm, usually the identity should be something like \(0\). More on this in a moment.
Also, notice that no matter where you are on the clock, there’s always an amount by which you can turn to end up at \(12\). For example, if you are at \(5\), you can add \(7\) to get to \(12\). This is called the . This number can easily be calculated by subtracting your number from \(12\). For example, \(12 - 5 = 7\) so the inverse of \(5\) is \(7\). Now what if we calculate the inverse of \(12\)? Then we get \(0\). In fact, \(12\) and \(0\) are the same thing on our clock. Actually, we may as well scratch the \(12\) off of our clock-face and just write \(0\). It makes things a lot easier. Now, instead of considering \(12\) the identity, we can consider \(0\) the identity instead. If we are at any number, and we step \(0\) times (in other words, we don’t step at all), then we stay at the number we were.
Notice now that instead of saying when we add \(5\) and \(7\) on our clock we get \(12\), we actually want to say we get \(0\). So now we change our rule at the top a little bit. To calculate the sum of two numbers, we add them, and if we get \(12\) or more we subtract \(12\). So now \(4\) plus \(2\) is still \(6\), and \(7\) plus \(9\) is still obtained by calculating \(7+9 = 16\) and \(16-12 = 4\). But now, instead of calculating that \(7 +5 = 12\) we now further calculate that \(12 - 12 = 0\) and so say that \(7\) plus \(5\) is \(0\) on our clock.
Now, there is no reason our clock needs to have \(12\) numbers. We could do exactly the same with, say, \(15\) numbers. Then our operation is defined by adding the numbers, and if they add up to at least \(15\), we subtract \(15\). Now \(5 \bullet 7 = 12\) but \(7 \bullet 9 = 1\) since \(7 + 9 = 16\) and \(16 - 15 = 1\). Notice that now the inverse of \(5\) is not \(7\), but is instead found by calculating \(15 - 5 = 10\). So the inverse of \(5\) is \(10\) (since \(5 + 10 = 15\), so \(5 \bullet 10 = 0\)).
We can write our inverses by adding a negative sign to the front of the number. So on our \(15\) number clock, the inverse of \(5\) is written \(-5\) and so \(-5 = 10\), but on our normal clock, the inverse of \(5\) is \(7\) so \(-5 = 7\).
This motivates the term cyclic; if we go far enough, we get back to where we started. In fact, a group is called cyclic if it has a singe \(1\). We can get any number by going \(1 \bullet 1 \bullet 1 \bullet \cdots \bullet 1\) the right number of times.. This is an element that can be used to generate all the other elements by being added to itself enough times. On any clock, the generator is
Technically, for mathematical reasons, we actually want to require that we can get every element even if we combine the generator with itself a negative number of times. What does this mean? Well, the generator \(1\) has an inverse \(-1\). In our normal clock \(-1 = 11\). In our \(15\) number clock \(-1 = 14\). The point is that we could just as well get all the numbers by adding \(-1\) to itself a bunch of times.
We can also get \(0\) by “not adding \(1\) to itself at all” or, put another way “adding \(1\) to itself \(0\) times”. This may seem strange, but if you think of taking \(1\), and then adding \(-1\) once, it’s like not taking \(1\) at all and you get \(0\). Don’t worry too much about this. The point is that you can technically get \(0\) from \(1\).
In fact, the isomorphic to some clock. That is to say, no matter what group you took that has this property, if you just renamed the elements, you would end up with something that looked like a clock in the way described above.that you can get from adding one number to itself a bunch of times are all
Take for example a coin. It has two sides: heads and tails. Call them \(h\) and \(t\) for short. We can define an operation on the sides as follows, which we’re going to refer to as “adding” sides: if you add heads to anything, you flip the coin. If you add tails to anything, you leave the coin alone. So, if you are on the heads side, and you “add heads”, you flip the coin to tails. If you are on the heads side and you “add tails” you leave the coin on the heads side. If you are on the tails side and you “add heads”, you flip the coin to heads, and if you are on the tails side and you “add tails” you just stay where you are. We can write the operation of ’adding” the sides as “$\bullet$” and then say \(h \bullet h = t\), and \(h \bullet t = h\), and \(t \bullet h = h\), and \(t \bullet t = t\). But hey! This is exactly the same we’d get if we called the head side \(1\) and the tails side \(0\), and considered a clock with just two numbers. Then \(1 \bullet 1 = 0\), and \(1 \bullet 0 = 1\), and \(0 \bullet 1 = 1\) and \(0 \bullet 0 = 0\). If you just had heads, you could get tails by just adding heads to itself, so the coin group is cyclic.
For any finite group where you can get every element starting from a single one, you can write it as a clock by relabeling the elements. Try it!
Infinite Cyclic Groups
However, things change if we allow our group to be \(1\) to itself some amount of times, either positive or negative or zero. But maybe there is no way to wrap around. Consider all the integers, that is to say, every number, positive, negative and zero. These actually do form a group. Then every number you can think of, you can get to by adding \(1\) a bunch of times. But no matter how many \(1\)s you add on the positive side, you can never wrap around and get a negative number. However, the integers are still called a cyclic group, even though it’s not really cyclic (I know, not a great name) because the official definition is that every element can be obtained by adding a specific something a bunch of times.. Then we still require that every element can be obtained by adding
Again, the only infinite cyclic group that you get, up to relabeling, is the integers.
So the only cyclic groups are the clocks and the integers!