# How many generators can a cyclic group have by definition?

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Hi, so I have just a small question about cyclic groups. Say I am trying to show that a group is cyclic. If I find that there is more than one element in that group that generates the whole group, is that fine? Essentially what I am asking is that can a cyclic group have more than one generator?

For example in the group ℤ14* = {1,2,5,9,11,13}, both 3 and 5 are generators. So is this ok to say that this group is cyclic?

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Yes, it is o.k. Do you have an idea, which property does make an element a generator?

Yes, it is o.k. Do you have an idea, which property does make an element a generator?
Yes, it is an element that when applied to itself with the relevant operation will generate the whole group that the element is a part of. Not the most eloquent way of putting it, but I understand it.

The reason I asked this question was because I was looking at this question here: https://i.gyazo.com/a377c9cf2e7c573a992d9fcbe1b1df37.png

Now, from what I understand, we have a rotation about a point on the line and a translation as our symmetries preserving the frieze. The rotation I believe is a generator since if you rotate once, it will give a rotation (one of the elements of the group of symmetries) and if you rotate once more, you will get back to a translation (the other element of the group of symmetries). Where this translation takes the frieze to depends on where we rotated about. So essentially I was wondering if there are infinite generators, such as rotation about point let's say (10,0), rotation about (0,0), etc. Is that correct to say? And if so, am I correct in saying it doesn't matter if there is more than 1 generator, and so the group is cyclic?

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I don't think your example is a cyclic group. You need two different elements to go back to the initial figure and neither is a power of the other. So at best, you can have a direct product of cyclic groups, but I haven't worked out the problem. One reason is that I don't know, whether the points indicate a sort of infinity and a translation is allowed because of it, since otherwise the translated figure doesn't coincide with the original.

Back to the cyclic groups. Yes, this ...
Yes, it is an element that when applied to itself with the relevant operation will generate the whole group that the element is a part of.
... is the definition. For an infinite cyclic group we get all ##m \cdot \mathbb{Z}## which are all isomorphic to ##\mathbb{Z}## and generated by ##\pm m##. So let's turn to the finite case. In this case we have a group ##G## generated by an element ##a## of say order ##n##. Then any element ##b## that also generates ##G## has to fulfill ##a=b^k## for some number ##k## and all elements have to be a power of ##a## as well as a power of ##b##. Now what will happen, if ##n## and ##k## are coprime, and what if they are not?

I don't think your example is a cyclic group. You need two different elements to go back to the initial figure and neither is a power of the other. So at best, you can have a direct product of cyclic groups, but I haven't worked out the problem. One reason is that I don't know, whether the points indicate a sort of infinity and a translation is allowed because of it, since otherwise the translated figure doesn't coincide with the original.

Back to the cyclic groups. Yes, this ...

... is the definition. For an infinite cyclic group we get all ##m \cdot \mathbb{Z}## which are all isomorphic to ##\mathbb{Z}## and generated by ##\pm m##. So let's turn to the finite case. In this case we have a group ##G## generated by an element ##a## of say order ##n##. Then any element ##b## that also generates ##G## has to fulfill ##a=b^k## for some number ##k## and all elements have to be a power of ##a## as well as a power of ##b##. Now what will happen, if ##n## and ##k## are coprime, and what if they are not?

Yes, the identity of that group is T0, which is a translation of 0 to the right, so it just stays in place. That has order 1. Every other translation has infinite order here however. So R02 gives us the rotation about (0,0) twice and so gives us the identity T0. Rn has order 2 here.

Is that not true that R0 generates the whole group? The group is just the 2 elements, the rotation and the translation. And it generates both.

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Yes, the identity of that group is T0, which is a translation of 0 to the right, so it just stays in place. That has order 1. Every other translation has infinite order here however.
Yes, this means a shift ##T_1^n## by ##n \in \mathbb{Z}## steps yields ##\langle T_1 \rangle \cong \mathbb{Z}## as a subgroup.
So R02 gives us the rotation about (0,0) twice and so gives us the identity T0. Rn has order 2 here.
Not sure which angle you attach to the rotation and what the index ##0## shall indicate here, as you used it for the identity element ##T_0##. So do you mean ##R_0 = 1##, i.e. no rotation at all, ##R_0 = R(\pi)##, i.e. a rotation by ##180°## and ##R_0^2=1## or ##R_0 = R(\frac{\pi}{2})##, i.e. a rotation by ##90°## and ##R_0^4 = 1## or even another angle.
Is that not true that R0 generates the whole group? The group is just the 2 elements, the rotation and the translation. And it generates both.
Say we have ##R_0=R(\frac{\pi}{2})## a rotation by ## \frac{\pi}{2} = 90°##. Then ##R_0## isn't part of the group, since it doesn't fix the figure. But all powers ##4n## of ##R_0## are, since ##R_0^{4n}=1##. In this case we have ##\langle R_0 \rangle \cong \mathbb{Z}/4\mathbb{Z} \cong \mathbb{Z}_4##, but the subgroup that leaves the figure invariant is ##\langle 4R_0 \rangle \cong 4\mathbb{Z}##, although the rotation itself is of order ##4##. And there are also two reflection axis, which gives transformations of order ##2##, and if applied twice leave the figure invariant.

So there is a difference between the transformations which leave the figure fixed, and their generators as an operation of the plane. If we don't count the many identities by always rotating full circles, there are only the translations left as true transformations.