Talk:Cyclic group

transl. de:Zyklische Gruppe

Did a nearly universal change in notation take place? Why do most people write $\mathbb {Z} /n\mathbb {Z}$ nowadays instead of $\mathbb {Z} _{n}$? Phys 17:02, 30 Aug 2003 (UTC)

Perhaps because Z_p is also used for the p-adic integers. Or perhaps because Z_n requires a special definition, whereas Z/nZ just reuses other standard notation. --Zundark 17:28, 30 Aug 2003 (UTC)

we could mention it's the quotient group notation in the article :) -- Tarquin 20:30, 30 Aug 2003 (UTC)

I've also seen the notation J_n for the finite cyclic groups (where of course J is used for the infinite cyclic group). This avoids the conflict with Z_p as the p-adic integers, but it appears this still leaves some ambiguity as J₁, J₂, J₃ and J₄ are commonly used for the Janko groups. Still, there is an advantage to using J_n or Z_n as it then makes sense to use the same notation for the corresponding finite rings, whereas C_n is unintuitive in that context. CyborgTosser 21:31, 28 Jun 2004 (UTC)

The remark about the discrete logarithm surely belongs in the final section, not with Z/nZ as additive group.

Charles Matthews 12:11, 18 Nov 2003 (UTC)

I always prefer Z/nZ for the cyclic group, rather than Z_n, because I work in number theory, and when n is prime, the latter means the ring of p-adic integers, not the cyclic group with p elements. The former notation is never ambiguous. Revolver

What is your definition of 'C_n'? This isn't a joke or a dumb question. You say ' for any positive integer n, there is a cyclic group C_n of order n ', but you never actually say precisely what the elements of C_n are. There are many different realisations of the cyclic group with n elements, so it almost seems as if you're using the C_n notation to denote the entire equivalence class of all groups isomorphic to a cyclic group with n elements. I know this probably sounds ridiculous, but there comes a problem when you say 'C_n is isomorphic Z/nZ', because how are you supposed to have an isomorphism when you haven't even told me what the elements of C_n are? This is something a lot of algebra textbooks do with respect to the cyclic groups, and it drives me nuts. In case you're wondering what possible solutions I have in mind, you could define C_n in terms of generators and relations (in essence, a presentation) or you could define it as the set of integers {0, 1, ..., n − 1} with operation given by remainder of the sum after dividing by n. Either one is independent of the construction as a quotient group of Z, so then it makes sense to say they're isomorphic. Revolver 21:37, 28 Jan 2004 (UTC)

Okay, I see you say that C_n is 'represented by the symmetries of the regular n-gon'. This is almost what I was asking for, except if this is how you want to define C_n, you should just SAY that C_n IS the (rigid motion!) symmetries of the regular n-gon, not that C_n is 'represented' by the motions.

In any case, I don't much care for the C_n notation, mainly for the confusion that we have here...the realisation as the quotient group of the integers is clearly a fixed representative in the isomorphism class, while 'C_n' gives no clue as to what C_n actually is. Revolver 21:45, 28 Jan 2004 (UTC)

Hi, me again...sorry to be a nuisance. I was looking around, I notice that a lot of articles use the C_n notation, esp. in the context of rigid motions symmetry groups and group presentations. So, it seems (imo) the best way to go is to either define C_n as the symmetries of the n-gon, then note that this is isomorphic to presentation ({x},{x^n}), or else to define it as the presentation and note this is isomorphic to symmetries. The same thing could be done for the dihedral groups. Which solution you like probably depends on whether you think of these groups as given by generators and relations, or as symmetries of geometries figures. But at least one choice needs to be made. Once a choice is made, you can show they're isomorphic and agree to use the same notation for each. But you can't slip in C_n in the back door without defining it, say that it's isomorphic to one of the 2 choices, then show they're both isomorphic. You need to start off by explicitly defining it as one or the other. Revolver 21:56, 28 Jan 2004 (UTC)

Isn't it best to define C_n as the quotient of 'the' free group on one generator, by its subgroup of index n? This impacts on the infinite cyclic group, so start with that?

Charles Matthews 17:02, 30 Jan 2004 (UTC)

That sounds like an acceptable way to do it, since free groups can be proved to be unique up to isomorphism. Is this any different from the group presentation? Revolver 19:50, 30 Jan 2004 (UTC)

Not in any fundamental way.

Charles Matthews 20:25, 30 Jan 2004 (UTC)

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In Examples of cyclic groups: The group of rotations in a circle, S1, is not a cyclic group.

My 2 questions are: Isnt this group named U(1)? And if it is not cyclic, so what is it then?

200.154.215.124 01:58, 1 Apr 2004 (UTC)

U(1,R) would be the group of all 1x1 orthogonal real matrices, i.e. the group {I, -I} consisting of just the identity matrix and its inverse.

Of course, now, (after morning coffee), I realise you must have meant U(1,C), 1-dim unitary group over C, and this is exactly S^1, or SO(2,R), you are correct. Revolver 22:50, 1 Apr 2004 (UTC)

Geometrically, this is the isometries of the line preserving the origin -- there are only two of them, -I representing the "flip" across the origin. I think the group you may be thinking of is SO(2,R), the special orthogonal group of all rotations of the plane fixing the origin. This group is isomorphic to S^1, the group mentioned in the article here, group of rotations of the circle, which is isomorphic to the multiplicative group of all complex numbers of absolute value one. (the unit circle) Multiplication of complex numbers corresponds to rotation in the unit circle, corresponds to addition of angle measure in the matrix representation (cos θ sin θ | −sin θ cos θ), and here is where we can get a fundamental description of S^1 -- since the trig functions have period 2π, S^1 is isomorphic to R under addition, modulo 2π, since the value of the modulus doesn't matter, we can finally say that S^1 is isomorphic to R/Z under addition. This group is uncountable, so it can't possibly be cyclic. Revolver 02:19, 1 Apr 2004 (UTC)