If you read a book on group theory they're going to give you all sorts of theorems and properties about cyclic groups and cyclic subgroups. And there are quite a few reasonably useful things that can be shown about cyclic groups with some relatively simple proofs. In all reality, though, none of that is extremely important. What is important is an understanding of what a cyclic group is and how they work. I'll repeat the definition here.
Definition: Cyclic
Let G be a group. Then G is cyclic if there exists an a in G such that for each g in G there is an integer, k, such that ak = g. In this case, we use the notation G = <a>.
Let G be a group. Then G is cyclic if there exists an a in G such that for each g in G there is an integer, k, such that ak = g. In this case, we use the notation G = <a>.
When I first introduced Z(n), I called them the cyclic groups. That is what they're called but I don't want that to be misleading. There do exists plenty of other cyclic groups and I'll give an example of another one.
Example: Another cyclic group
I will use the standard notation of i2 = -1. Let G = {1, i, -1, -i} and let the operation on G be standard multiplication. It is easy to verify that G is a group. Now notice that i0 = 1, i1 = i, i2 = -1, and i3 = -i. It then follows from the definition of a cyclic group that G is cyclic and G = <i>.
I will use the standard notation of i2 = -1. Let G = {1, i, -1, -i} and let the operation on G be standard multiplication. It is easy to verify that G is a group. Now notice that i0 = 1, i1 = i, i2 = -1, and i3 = -i. It then follows from the definition of a cyclic group that G is cyclic and G = <i>.
We can generalize this notation of <a>, and this is often where we get the most usefulness out of this concept of cyclic groups. Suppose that G is a group and g∈G. We define <g> = {gk : k∈Z} = {...,g-2,g-1,g0,g1,g2,...}. In regular english, this means that <g> is the set of all elements of the form gk such that k is any integer. In case it is not clear, if n is positive, then g-n = (g-1)n = (gn)-1. We now prove that if g∈G then <g> is a subgroup of G.
Theorem: <g> is a subgroup
Let G be a group and g∈G. Then <g> is a subgroup of G and is called the cyclic subgroup generated by g.
Let G be a group and g∈G. Then <g> is a subgroup of G and is called the cyclic subgroup generated by g.
Proof:
First we must show that <g> is a subset of G. Choose an element gk∈<g> where k is an integer. If k = 0 then gk = e, the identity element of G which is clearly in G. If k ≠ 0 then gk is either the product of k copies of g (if k is positive) or the product of k copies of g-1 (if k is negative). But g,g-1∈G and G is closed under its operation, so gk∈G and <g>⊆G. I now wish to use the one-step subgroup test to show that <g> is a subgroup of G. Choose gm,gn∈<g>. Then gm∙(gn)-1 = gm∙g-n = gm-n and m-n is an integer so gm∙(gn)-1=gm-n∈<g>. Thus, by the one-step subgroup test, <g> is a subgroup of G.
First we must show that <g> is a subset of G. Choose an element gk∈<g> where k is an integer. If k = 0 then gk = e, the identity element of G which is clearly in G. If k ≠ 0 then gk is either the product of k copies of g (if k is positive) or the product of k copies of g-1 (if k is negative). But g,g-1∈G and G is closed under its operation, so gk∈G and <g>⊆G. I now wish to use the one-step subgroup test to show that <g> is a subgroup of G. Choose gm,gn∈<g>. Then gm∙(gn)-1 = gm∙g-n = gm-n and m-n is an integer so gm∙(gn)-1=gm-n∈<g>. Thus, by the one-step subgroup test, <g> is a subgroup of G.
I know that's kind of a dense proof and might be a little confusing, but the proof isn't that important or groundbreaking. Lets look at an example of one of these cyclic subgroups. Consider Z(6) and 2∈Z(6). I now wish to look at <2>. Every element of <2> is of the form k∙2 for each integer, k and I'd now like to investigate what this set looks like. <2> = {0∙2,1∙2,2∙2,3∙2,4∙2,...} = {0,2,4,0,2,...}. This sequence will continue indefinitely, so <2> = {0,2,4}. Similarly, <3> = {0,3}, and each subsequent cyclic subgroup can be calculated similarly.
Now, if G is a cyclic group, then we have just proven that ∀g∈G, <g> is a subgroup of G. But, by the definition of cyclic groups, we know that there exists some a∈G such that <a> = G. The question then arises as to whether this a is unique. For example, we know that Z(6) = <1> under addition mod 6, but is there any reason there isn't some other k∈Z(6) such that Z(6) = <k>? The answer is no. In fact, Z(6) = <5>. Notice that <5> = {0∙5,1∙5,2∙5,3∙5,4∙5,5∙5,6∙5,7∙5,...} = {0,5,4,3,2,1,0,5,...}. We know that there cannot be anything in <5> that is not in Z(6) (because <5>⊆Z(6)) but we've seen that each element in Z(6) is also in <5>, so <5>=Z(6). This brings us to the following definition.
Definition: Generator of a Cyclic Group
Suppose that G is a cyclic group and a∈G. Then if <a> = G, a is called a generator of G.
Suppose that G is a cyclic group and a∈G. Then if <a> = G, a is called a generator of G.
In general, most cyclic groups have more than one generator. There is one last thing that I need to say about cyclic subgroups. The examples of cyclic subgroups that I gave were subgroups of groups that were, themselves, cyclic. That was merely a coincidence. Given any arbitrary group, G, and any element a∈G, <a> is a cyclic subgroup of G regardless of the properties of the original group, G.
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