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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.

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

1 Answer 1

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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 $G\subset\langle x^a\rangle$,

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    $\begingroup$ Wow...you're a wizard. But don't you have to prove both ways? ie. G⊂(x^a) and (x^a) ⊂ G? $\endgroup$
    – koifish
    Jul 25, 2018 at 13:26
  • $\begingroup$ I suppose that your talkang about the $\subset$ sign. Yes, they're equal. The other inclusion is obvious. $\endgroup$ Jul 25, 2018 at 13:27
  • $\begingroup$ I see I see. Thanks! $\endgroup$
    – koifish
    Jul 25, 2018 at 13:28
  • $\begingroup$ @YipJungHon Any reason to mark my answer as accepted and then to un-mark it? Just curious. $\endgroup$ Jul 25, 2018 at 14:24
  • $\begingroup$ Nah, don't take it personally. I realized that after marking your answer I wasn't getting any other replies, so I was wondering if StackExchange was not showing a resolved question to the wider community. $\endgroup$
    – koifish
    Jul 25, 2018 at 14:47

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