# Cyclic and non-cyclic groups

October 31, 2006 8:46 AM Subscribe

If a group is said to be cyclic if it is generated by a single element (single generator), is the quaternion group an example of a finite group that is not cyclic, since it has four generators {-1, i, j, k}? Also, why is the group of rationals under addition not cyclic?

Your assertion that the quaternion group has four generators is faulty.

i*i = -1

i*j = k

Can you get j from {i, -1}? If so it's cyclic and if not it's dicyclic.

posted by unSane at 9:25 AM on October 31, 2006

i*i = -1

i*j = k

Can you get j from {i, -1}? If so it's cyclic and if not it's dicyclic.

posted by unSane at 9:25 AM on October 31, 2006

If I'm not mistaken it has to do with the rationals being a dense set of the real line. But it's been a while and my memory is foggy

posted by jplank at 9:49 AM on October 31, 2006

posted by jplank at 9:49 AM on October 31, 2006

suppose there was some rational number r that generated the rational numbers

so the rational numbers can be expressed as

{0, r, r+r, r+r+r... } = {0, r, 2r, 3r ...}

in other words, every rational number can be expressed as n*r where n is some integer

now think about r/2

posted by unSane at 10:08 AM on October 31, 2006

so the rational numbers can be expressed as

{0, r, r+r, r+r+r... } = {0, r, 2r, 3r ...}

in other words, every rational number can be expressed as n*r where n is some integer

now think about r/2

posted by unSane at 10:08 AM on October 31, 2006

*{0, r, r+r, r+r+r... } = {0, r, 2r, 3r ...}*

now think about r/2

now think about r/2

This is right, mostly, but don't forget inverses! If you've got {0, r, 2r, 3r ...} then you've also got {-r, -2r, -3r ...}

The quaternions can indeed be generated by the four elements you list. But they can also be generated by just {1, i, j}, since ij=k and you get inverses for free. To say that Q is

*not*cyclic means you can't get by with just one. Hint: if a group of order n is cyclic, there has to be an element of order n.

posted by gleuschk at 10:48 AM on October 31, 2006

Response by poster:

Got it. I think I am confusing number of generators with what

In the case of Q\{0} I guess it is about not be able to generate, for example, 5/7 from any additive multiple of, for example, 2/3 or its inverse.

posted by Mr. Six at 11:04 AM on October 31, 2006

*To say that Q is not cyclic means you can't get by with just one.*Got it. I think I am confusing number of generators with what

*multiple*of a specific generator is required to make it cyclic.In the case of Q\{0} I guess it is about not be able to generate, for example, 5/7 from any additive multiple of, for example, 2/3 or its inverse.

posted by Mr. Six at 11:04 AM on October 31, 2006

Every finite cyclic group of order n is isomorphic to the integers 0 through n-1 under modulo n addition. The quaternion group (order 8) is cyclic iff it is isomorphic to the integers 0-7 under modulo 8 addition. Whether it is or not is left as an exercise to the reader.

posted by DevilsAdvocate at 11:19 AM on October 31, 2006

posted by DevilsAdvocate at 11:19 AM on October 31, 2006

One source of confusion with the "are rationals cyclic?" issue is that the integers and rationals are of the same cardinality. So, yes, there are functions which allow you to do a 1-1 mapping of integers and rationals (i.e. show they are isomorphic) - its just not a simple multiplicative function.

posted by vacapinta at 11:24 AM on October 31, 2006

posted by vacapinta at 11:24 AM on October 31, 2006

*The quaternions can indeed be generated by the four elements you list. But they can also be generated by just {1, i, j}, since ij=k and you get inverses for free.*

Actually they can be generated but just {i, j} as I tried to hint in my post:

i*i = -1

i*i*i*i = 1

i*j = k

posted by unSane at 12:11 PM on October 31, 2006

I think your confusion is represented by the statement "a group is said to be cyclic". "Cyclic" is not a quality like abelian. A Cyclic Group of order N is not a *type* of group, it *is* a group.

There is only 1 cyclic group of a given order. For a cyclic group of order N, there exists an element of order N. The subgroup generated by that element is the entire group. I think of it in terms of orbits. In a cyclic group, there is only 1 orbit: 1, x, x^2, ... x^n-1, and this orbit is the whole group.

therefore, conceptually, to determine if a group is cyclic or not isn't a matter of satisfying properties; it's whether or not the group is isomorphic to the cyclic group of order N (where N is the size of the group). If there are elements in different orbits, then it's not cyclic. If not, then the orbit is the entire group.

If the rationals under addition is not cyclic, it may be because cyclic groups must be finite. I can't recall if this is true or not, but I have a hunch.

posted by cotterpin at 1:05 PM on October 31, 2006

There is only 1 cyclic group of a given order. For a cyclic group of order N, there exists an element of order N. The subgroup generated by that element is the entire group. I think of it in terms of orbits. In a cyclic group, there is only 1 orbit: 1, x, x^2, ... x^n-1, and this orbit is the whole group.

therefore, conceptually, to determine if a group is cyclic or not isn't a matter of satisfying properties; it's whether or not the group is isomorphic to the cyclic group of order N (where N is the size of the group). If there are elements in different orbits, then it's not cyclic. If not, then the orbit is the entire group.

If the rationals under addition is not cyclic, it may be because cyclic groups must be finite. I can't recall if this is true or not, but I have a hunch.

posted by cotterpin at 1:05 PM on October 31, 2006

Nope, cyclic groups don't have to be finite. An infinite cyclic group is isomorphic to the integers under addition.

The name is confusing because it doesn't contain any cycles...

posted by unSane at 2:23 PM on October 31, 2006

The name is confusing because it doesn't contain any cycles...

posted by unSane at 2:23 PM on October 31, 2006

The rationals are not a cyclic group because they cannot be generated by a single element, nor can they even be

As justification, consider any rational number, p/q, where p is relatively prime to p. The group generated by p/q is all numbers of the form (p/q)^n, where n is an integer. In the world of rational numbers, this is a relatively small set, and certainly not all integers. For example, you are missing (p+1)/q.

I don't have my algebra texts handy at the moment, so I can't verify that this is how the name arose, but I believe the name captures how the group acts on sets and is not a reference to finite cycles existing in a group.

I'm feeling a bit lazy, so I'm going to link the wikipedia article on cyclic groups which has a good exposition of what I'm getting at in the 'representation' section. If you extend the action to an infinite set, it's straight forward to see how Z arises in this way.

posted by lastyearsfad at 3:14 PM on October 31, 2006

*finitely*generated as a group.As justification, consider any rational number, p/q, where p is relatively prime to p. The group generated by p/q is all numbers of the form (p/q)^n, where n is an integer. In the world of rational numbers, this is a relatively small set, and certainly not all integers. For example, you are missing (p+1)/q.

*The name is confusing because it doesn't contain any cycles...*I don't have my algebra texts handy at the moment, so I can't verify that this is how the name arose, but I believe the name captures how the group acts on sets and is not a reference to finite cycles existing in a group.

I'm feeling a bit lazy, so I'm going to link the wikipedia article on cyclic groups which has a good exposition of what I'm getting at in the 'representation' section. If you extend the action to an infinite set, it's straight forward to see how Z arises in this way.

posted by lastyearsfad at 3:14 PM on October 31, 2006

*p is relatively prime to p*

p is relatively prime to q....

posted by lastyearsfad at 3:15 PM on October 31, 2006

*As justification, consider any rational number, p/q, where p is relatively prime to p. The group generated by p/q is all numbers of the form (p/q)^n, where n is an integer. In the world of rational numbers, this is a relatively small set, and certainly not all integers. For example, you are missing (p+1)/q.*

The question is about rationals under addition, not multiplication. See my post above for the reductio ad absurdum.

posted by unSane at 4:53 PM on October 31, 2006

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arecyclic, and try to generate them from a single rational number?posted by Humanzee at 9:05 AM on October 31, 2006