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Cyclic groups are defined where repeated application of the group operation is applied, to generate whole group, to generator element.

But, have confusion as have seen questions that ask about generators of group for a given value of $\mathbb{Z_{n=p}}, \mathbb{Z_{n= p^r}}$ here.

As $+\pmod n$ is operation for Cyclic Group $C_n$, so shouldn't all elements of $C_n$ be generators, irrespective of whether $n$ is prime or composite?

$ \begin{array}{c|ccccc} + \pmod 6 & 0 &1 & 2& 3 & 4 & 5\\ \hline 0 & 0 & 1 &2 & 3& 4 & 5\\ 1 & 1& 2 & 3& 4 & 5 & 0\\ 2 & 2 & 3 & 4 & 5 & 0 & 1 \\ 3 & 3 & 4& 5& 0& 2 &0\\ 4 & 4 & 5& 0& 1& 2&3 \\ 5 & 5 & 0& 1& 2& 3& 4\\ \end{array} $

The need for totient function should instead arise only in multiplicative table $<\langle \mathbb{Z}/p\mathbb{Z} \setminus\{0\}, *\rangle$, but that doesn't form a group even.

$ \begin{array}{c|ccccc} \times \pmod 6 &1 & 2& 3 & 4 & 5\\ \hline 1 & 1& 2 & 3& 4 & 5 \\ 2 & 2 & 4 & 0 & 2 & 4\\ 3 & 3 & 0& 3& 0& 3\\ 4 & 4 & 2& 0& 4& 2 \\ 5 & 5 & 4& 3& 2& 1\\ \end{array} $

Here, only elements ($1,5$) given by totient function are generators.

Edit. the way operations are applied is not in the sense of cyclic.

The correct approach is: $\begin{array}{c|ccccc} + \pmod 6 & a^0 & a^1 &a^2 & a^3& a^4 & a^5\\ \hline 0 & 0 &0 & 0& 0 & 0 & 0\\ 1 & 0& 1 & 2& 3 & 4 & 5\\ 2 & 0& 2 & 4 & 0 & 2& 4 \\ 3 & 0 & 3& 0& 3& 0&3\\ 4 & 0 & 4& 2& 0& 4&2 \\ 5 & 0 & 5& 4& 3& 2& 1\\ \end{array} $

$ \begin{array}{c|ccccc} \times \pmod 6 &a^1 & a^2& a^3 & a^4 & a^5\\ \hline 1 & 1& 1 & 1& 1 & 1 \\ 2 & 2 & 4 & 2 & 4 & 2\\ 3 & 3 & 3& 3& 3& 3\\ 4 & 4 & 4& 4& 4& 4 \\ 5 & 5 & 1& 5& 1& 5\\ \end{array}$

Have still one more doubt: is the first column of the additive group having entries $=0$ correctly?

jiten
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    "Shouldn't all elements be generators?" No. The identity never is. Why don't you try an $n$ that isn't prime? Is $2$ a generator of the integers modulo $4$ under addition? And your second table is abject nonsense. The nonzero residues modulo $6$ do not form a group, as is patently obvious from the table you post. – Arturo Magidin Jul 01 '22 at 02:26
  • @ArturoMagidin $2$ for integers modulo $4$ under addition yields: $$2+0\equiv 2, 2+1\equiv 3, 2+2\equiv 0, 2+3\equiv 1.$$ So, why it not generates the group? – jiten Jul 01 '22 at 02:37
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    The multiplicative group of integers modulo $n$ has $\varphi(n)$ elements by definition: its elements are precisely the remainders represented by integers relatively primeto $n$. – Arturo Magidin Jul 01 '22 at 02:38
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    No, generator means using only that element. If you are so utterly confused about what things mean, why do you insist on wasting your time and everyone else's by not learning them properly? With $2$ mod $4$ you only get $2$, $2+2=0$, and nothing else. Why would you even think you get to add $1$? Nonsense. – Arturo Magidin Jul 01 '22 at 02:41
  • @ArturoMagidin Do you mean the way am applying operations is wrong? Then, correct answer is: $$0, 2, 0, 2$$ – jiten Jul 01 '22 at 02:43
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    I mean that you obviously do not understand what "generator" means, yet you try to figure out what is and what is not a generator. As you clearly do not know, nor care to spend your time finding out (instead demanding people feed you pap in the form of links) then you are, naturally, confused. The definitions are not hard, and easy to find. So make the effort yourself. – Arturo Magidin Jul 01 '22 at 02:47
  • @ArturoMagidin kindly see the edit. – jiten Jul 01 '22 at 02:55
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    Please take my name off your misunderstandings. I do not wish anyone to think I have anything to do with your nonsense. – Arturo Magidin Jul 01 '22 at 03:00
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    You're swimming in details you don't understand, and it makes it really difficult to straighten you out. Maybe you can eventually get your footing. But it's going to take a while. (Some of) it's par for the course. High-powered mathematics can be deceptively difficult. – suckling pig Jul 01 '22 at 03:48

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First, I am your downvoter. You keep biting off more than you can chew. It's quite a mess.

However, here's one more college try:

A basic fact about cyclic groups, which it would be great progress if you would learn to prove, is that $|g^k|=n/\gcd(n,k)$, where $n=|g|$.

This explains why Euler's totient function comes into play.

suckling pig
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  • Isn't this Lagrange theorem? Though the proof of your theorem seems more algebraic than based on cosets, as in Lagrange theorem. I said algebraic though am not clear about proof. Though, intuitively it is must for $g$ to cover group. But, still need handle proof of subcases, and so stuck. – jiten Jul 02 '22 at 10:26
  • Seems not on that topic even remotely. – jiten Jul 02 '22 at 23:28
  • I'm sorry, somehow it was the wrong link. Here https://math.stackexchange.com/a/4484127/1070376 – suckling pig Jul 02 '22 at 23:33