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I am trying to provide a solution to the following exercise. Please point out anything that you find wrong and/or bad.

Show that a cyclic group $G$ of order $n$ generated by an element $a$ can also be generated by $a^k$ if $k$ and $p$ are relatively prime.

MY ATTEMPT:

Any element of $G$ can be written as

$$a^x,\,1\leq x\leq n$$

We are trying to show the following

$$\left(a^k\right)^m=a^x\,\,\, (\textrm{mod}\,\,n)$$

where $k$ and $n$ are relatively prime, and $m$ is some integer less than $n$. By Bézout's Identity,

$$(k,n)=1\implies ak+bn=1$$

$$\begin{align*} \rightarrow (xa)k+(xb)n&=x\\ (m)k+(\beta)n&=x\\ (m)k&=(-\beta)n+x\\ mk&=\alpha n +x\\ &\\ \implies mk&=x\,\,\,(\textrm{mod}\,\,n) \end{align*} $$

We have found our integer $m$ that satisfies the equality. It is left to show that each $(a^k)^m$ for $1\leq m \leq n$ is a distinct member of $G$. Note that if $m\geq n$, then $m=qn+b$ for some positive integers $q$ and $b$, where $1\leq b \leq n$. This means that

$$\left(a^k\right)^m=\left(a^k\right)^{qn+b}=a^{(kq)n}a^{kb}=(e)a^{kb}=\left(a^k\right)^b$$

And so the situation is reduced back to the original one. Thus, because $(a^k)^m$ covers every element in $G$ for $1\leq m\leq n$, there is a bijection between the group generated by $a^k$ and $G$. $\blacksquare$.

Arturo don Juan
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  • It's a fine proof. Just the fact that all $(a^k)^m$ are distinct. Not sure it's necessary, as each $a^x$ is attained (don't forget for finite sets with the same cardinal, surjective implies injective). – Bernard Feb 14 '15 at 23:22
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    This question is similar to: Cyclic Group Generators of Order $n$. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Dean Miller Dec 31 '25 at 05:46
  • @DeanMiller: Since this post is two years older than the one you found, and since this one has an accepted answer, would it be better to close that later one? – Lee Mosher Dec 31 '25 at 20:37
  • @LeeMosher The question there has higher quality answers and is also answering a more broad question, namely what are the generators of a finite cyclic group and how many are there, so I think that one should definitely be left open. This one also has only one upvote (not much interest) and essentially asks us to check a proof without any indication of what is in doubt, so perhaps it can be closed as needing details or clarity. But I strongly suggest the other question should be reopened regardless of what happens to this one due to its quality and that it answers a more general question. – Dean Miller Dec 31 '25 at 22:12

2 Answers2

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Essentially correct.

Faster: Note that since $(k, n)=1$, $\exists p,q$ such that $pk + qn = 1$. But then $$ a^{pk + qn} = a \\ \implies a^{pk} \cdot a^{qn} = a $$ but $a^{qn} = 1$, so $$ a^{pk} = a, $$ and $a$ generates the cyclic group, so $a^k$ does too.

Michael Cromer
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Hint $\ a^k\,$ has order $\,\color{#c00}{n/(k,n)}\,$ by basic gcd laws (below). $ $ But $\,n/(k,n) = n\iff (k,n)= 1.$

$$a^{kj} = 1 \iff n\mid kj \iff n\mid kj,nj \iff n\mid (kj,nj) = (k,n)j \iff \color{#c00}{n/(k,n)}\mid j\qquad $$

Bill Dubuque
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