# Group Theory: Proving if $G = \langle x \rangle$, $|G| = r$, then $G' = \langle x^a \rangle = G$ provided $a$ is coprime to $r$.

So this is my question:

Let $$G$$ be the cyclic group generated by the element $$x$$, i.e., $$G = \langle x \rangle = \{ 1, x, x^2, x^3, \dotsc, x^{r-1} \}$$. Now, $$|G| = r$$. Prove that $$G' = \langle x^a \rangle = G$$ if $$a$$ and $$r$$ are coprime (i.e., the cyclic group generated by $$x^a$$ is equal to the cyclic group generated by $$x$$).

I get the intuition behind it. For instance, if $$a$$ and $$r$$ wasn’t coprime, the subgroup generated by $$x^a$$ would not generate any new elements once it goes back to the identity.

On the other hand, if $$a$$ and $$r$$ was coprime (e.g., $$r = 30$$ and $$a = 7$$), then $$\langle x^7 \rangle$$ would be able to generate every single element of the group.

This is a lemma that I generated for myself to solve this question. I’m not sure if this is a well-established proof in math and I couldn’t find any results on Google.

Lemma: Two coprime numbers do not divide.

Let $$a$$ and $$b$$ be coprime. $$a$$, $$b$$ can be expressed as a product of some prime factorization. $$a = p(1)^{a(1)} \cdot p(2)^{a(2)} \dotsm \,,\quad b = p'(1)^{a'(1)} \cdot p'(2)^{a'(2)} \dotsm$$ $$p(x)$$ does not equal $$p'(x)$$ as if it does, it would mean that we have found a factor $$p(x)$$ which divides both numbers. Because all primes used in the prime factorization of $$a$$ and $$b$$ are different, $$a$$ does not divide $$b$$ and vice versa.

The problem is that I don’t know how to prove the above question vigorously. I was thinking of using the Lagrange’s theorem, and arguing that $$|\langle x \rangle|$$ must divide $$|\langle x^a \rangle|$$. So if $$a$$ and $$r$$ weren’t coprime, then $$|\langle x \rangle|$$ would divide $$|\langle x^a \rangle|$$ and so $$\langle x^a \rangle$$ would be a subgroup of $$\langle x \rangle$$.

On the other hand, if $$a$$ and $$r$$ were coprime, then if we’d assume that $$\langle x^a \rangle$$ cannot contain all elements of $$\langle x \rangle$$, then $$\langle x^a \rangle$$ would be a proper subgroup of $$\langle x \rangle$$. If this subgroup exists, let’s call its order $$r'$$. But $$r$$ does not divide $$r'$$! So we have a contradiction and thus $$\langle x^a \rangle$$ cannot be a proper subgroup of $$\langle x \rangle$$. Thus $$\langle x^a \rangle = \langle x \rangle$$, i.e. $$G' = G$$!

I’m not sure if I’m expressing myself very clearly here. Basically, I’m trying to show if you pick an arbitrary element from a cyclic group, and generate the group from it, you will either get a subgroup of the cyclic group or the entire group again. For $$a$$ is coprime to $$r$$, there is no way in hell you’re going to get a proper subgroup of order $$a$$ as I’ve established that $$r$$ does not divide $$a$$. Thus $$x^a$$ must generate the entire group.

If there is a simpler way to prove it, please let me know. And also, I would be grateful if someone with a solid understanding of group theory critiques my proof.

• That coprime numbers do not divide, is kinda the definition of beeing coprime. Jul 25, 2018 at 13:19
• Well...haha thanks! I'm not that familiar with the concept of coprime numbers. Jul 25, 2018 at 13:22

SInce $r$ and $a$ are coprime, Bézout's lemma tells us that there are integers $m$ and $n$ such that $rm+an=1$. So$$x=x^1=x^{rm+an}=(x^a)^n.$$This proves that $x\in\langle x^a\rangle$ and that therefore $G\subset\langle x^a\rangle$,
• I suppose that your talkang about the $\subset$ sign. Yes, they're equal. The other inclusion is obvious. Jul 25, 2018 at 13:27