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This is a question that was asked in my group theory examination today:

Let $G$ be a finite cyclic group generated by an element $x$ of $G$. If $y(\ne x)\in G$ is also a generator of $G$, find the relation between the elements $x$ and $y$.

I do not think that given an arbitrary finite cyclic group one can give a nice relation between any two of its generators. For example if $\mathbb{Z}/50\mathbb{Z}$ what is the relation between 7 and 49 or 23 and 31 or say 3 and 43? I have not been able to understand clearly what kind of a relation the question asks for. I know that $x=y^m$ for some $m$, and $m$ is then coprime to the order of the group but I do not know how could this give a relation involving only $x$ and $y$. So what is the question asking for and what in general is a relation between $x$ and $y$?

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up vote 3 down vote accepted

Well, there are some very basic relations between $\,x\,,\,y\,$, for example say $\,|G|=n\,$ , then:

$$\begin{align*}*&\;\;\;|x|=|y|\\**&\;\;\;y=x^k\,\,,\,\,x=y^s\;\;,\;\text{for some}\;\;k,s\in\Bbb Z\;\;\text{with}\;\;(n,k)=(n,s)=1\end{align*}$$

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The first relation you give could be a correct answer I suppose since that is a relation between $x$ and $y$(and that is the only relation that I can think of). But I wrote the second one as an answer. Thanks for it anyways. – A. Gupta Sep 23 '12 at 9:42
@A.Gupta: Was this a take-home examination? – user1729 Sep 23 '12 at 13:55
@ user1729 -No, this was not a take home examination. Take home examinations are not as common(not in my college atleast) in India as in other places like US or UK(as far as I know). But, this was an 'open-notes' exam although I didn't carry my notes to the exam hall. Have a nice day! – A. Gupta Sep 23 '12 at 15:52
@A.Gupta: Sorry, it was just the wasy you said "...I wrote the second one as an answer" which made me wonder if you had the exam in front of you and were just copying the answer! (Which would be...wrong?) – user1729 Sep 24 '12 at 15:01

Your answer sounds fairly reasonable. If $y$ is also a generator then the order of $y$ is the same as the order of $x$. Since $G$ is cyclic $y=x^k$ for some $k<|G|$ but $y^m=e$ since it generates $G$. This will indicate $k$ is relatively prime to $m$. Towards a contradiction suppose otherwise, suppose $m=kl$ then $y^l=(x^{k})^l=x^{kl}=x^m=e$ which says $|y|=l \neq m$. Hence $m$ and $k$ are relatively prime.

As an example, if $G = <c>$ for $|G|=10$ then $c,c^3,c^7,c^9$ are the possible generators. I would wager your professor was fishing for something like what you say in your post.

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Thanks for the explanation. I also think something similar was asked for. – A. Gupta Sep 23 '12 at 9:43

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