Is only one generator enough to find other generators?

Question

I am self learner high school student interested in abstract algebra,nowadays.

I saw the followings in my book: "If a is a generator of a finite cyclic group G of order n, then the other generators of G are the elements of the form $a^r$ , where r is relatively prime to n." I wonder whether we count all generators by finding all elements in the form of $a^m$ where $(m,n)=1$ by using a single generator $a$, because if we have another generator call $b$, is there any other generator in the form of $b^k$ where $(k,n)=1$ . Is only one generator is enough to find all other generators ?

Note: I read the proof of foregoing theorem, but I could not conclude whether a single generator is enough to produce all other generators. Can you please enlighten me by thinking my situation. A proof can be good for me to read it while I study in my free time.

I am sorry if this question is not good for this site,but I could not ask it to my teachers, so I have appealed here.

Answer 1

As Sean Eberhard has pointed out, the powers of a generator $a$ give you all the elements of the group. Thus the only thing left to prove is that if $(m, n) = d > 1$, then $b = a^m$ is not a generator.

Suppose $b$ is a generator. Then for some $k$ we have $b^k = a$, hence $1 = a^{n(m/d)} = b^{n/d}$, so there cannot be more than $n/d < n$ distinct powers of $b$ and hence $b$ is not a generator.

Answer 2

No need to apologize, your question is perfectly suited for this forum and is an interesting topic in abstract algebra! Understanding the role of generators in a cyclic group is quite important.

The easy answer to your question is: yes, a single generator is enough to generate all other generators of a finite cyclic group G of order n.

To understand why, let's start with something important:

A group G is called cyclic if there exists an element a (the generator) in G such that every element of G can be written as powers (i.e., repeated applications of the group operation) of a.

Now, in the context of your specific question, assume that G is a finite cyclic group of order n generated by the element a. If we select some integer r that is relatively prime to n (meaning the greatest common divisor (gcd) between r and n is 1), then a^r is also a generator of G. Conversely, every generator of G is in the form of a^r where gcd(r, n) = 1.

So, now you might wonder, if we find another generator let's say b = a^r where gcd(r,n) = 1, does there exist some s where gcd(s, n) = 1 such that b^s = a? The answer is yes!

Consider s - the multiplicative inverse of r modulo n. Because r and n are relatively prime, an inverse exists. This means rs ≡ 1 (mod n). Hence:

b^s = (a^r)^s = a^(rs), which by the multplicative inverse relationship is a

Therefore, yes, you can indeed generate all generators of G using just one generator.

In a practical sense, this is really handy because it means you can explore and understand the entire structure of a cyclic group by simply manipulating a single chosen generator.

This mathematical property holds because of Euler's theorem for finite cyclic groups, which has some nice and intuitive proofs. It's actually a very special property of cyclic groups and doesn't hold for groups in general, which makes cyclic groups that much more interesting to study.

Hopefully this helps!

Answer 3

Perhaps the statement you are missing is this, which is just a rewording of the statement you have quoted: For all $b \in G$, $b$ is a generator of $G$ if and only if there exists an integer $r$ relatively prime to $n$ such that $b=a^r$.

So once you have found all those integers $r$, and computed all those powers $a^r$, this statement guarantees that you have found all other generators $b$.

Now, you seem to be concerned about this situation: once having found another generator $b=a^r$, the theorem also tells you that all these other powers $b^k$ with $k$ relative prime to $n$, are also generators. However, since $b^k = (a^r)^k = a^{rk}$, and since both $k$ and $r$ are relatively prime to $n$, you can use a little bit of calculation to convince yourself that $rk$ is also relatively prime to $n$. So, you've already found $b^k$! It's nothing new, it's equal to $a^{rk}$.