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 in­verse. 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.

Our clock is closed un­der its ad­di­tion, it has an iden­tity el­e­ment and in­verses. Hence, our clock is a group. From here on out we can write our group op­er­a­tion as \(\bullet\). So \(7 \bullet 9 = 4\).

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 \(-5 = 7\).

Cyclic Groups

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 gen­er­a­tor. 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 \(1\). We can get any num­ber by go­ing \(1 \bullet 1 \bullet 1 \bullet \cdots \bullet 1\) the right num­ber of times.

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 finite groups that you can get from adding one num­ber to it­self a bunch of times are all 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.

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 in­finite. Then we still re­quire that ev­ery el­e­ment can be ob­tained by adding \(1\) 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.

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!


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