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A cyclic group is a group that can be generated by repeatedly applying the group operation to an element of the group called the "generator".

I noticed that, at least for the multiplicative group of integers modulo $5$, the equations $x = 2^y \pmod{5}$ and $x = 3^y \pmod{5}$, where $x$ is any element of said group, both have infinite solutions. $2$ and $3$ are generators of this group.

Ascending powers of either generator modulo 5 produce a unique, endless pattern. For example:

$$2^1\pmod{5} = 2$$

$$2^2\pmod{5}= 4$$

$$2^3\pmod{5} = 3$$

$$2^4\pmod{5} = 1$$

$$2^5\pmod{5} = 2$$

$$2^6\pmod{5}= 4$$

$$2^7\pmod{5}= 3$$

The repeating pattern contains all the elements of the group.

Is there a proof that any cyclic group generator produces an endlessly repeating pattern containing all elements of its group?

EDIT: I just realized what I'm asking is answered here: Why does a number like $2^n\bmod 7$ have a repeating pattern?

user363406
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  • Yes. You can prove it with long division in the exponent. It's also obvious because you have an infinite number of exponents but only finitely many group elements. – Randall Nov 12 '21 at 19:32
  • In finite groups, as the tag says though not the question, the pigeonhole principle guarantees this, otherwise you would run out of elements to assign $a^n$ to – Alan Nov 12 '21 at 19:34
  • Note also that $5$ is a prime and there is always a primitive root for a prime $p$ ie a number whose powers will cycle all the non-zero integers less than $p$. Not every number does this, and the existence of a primitive root is equivalent to saying that the multiplicative group of residues modulo a prime is always cyclic. – Mark Bennet Nov 12 '21 at 20:31

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Yes.

For all generators $a$ of a cyclic group $G$, for each $g\in G$, there exists an $n\in\Bbb N$ such that $a^n=g$ by definition of $G$ being cyclic; that includes letting $g$ be another generator. For finite groups, it follows from the Pigeonhole Principle. For the only infinite cyclic group, $(\Bbb Z, +)$, up to isomorphism, the generator is either $1$ or $-1$, each of which generates each element as a multiple.

Shaun
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