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Suppose that $d$ and $k$ are positive integers satisfying $dk = n$. Show that there exists a unique subgroup of $Z_n$ of order $d$ generated by $[k].$

Then, suppose that the cyclic group $G$ operates on a set $S$ and $g_1$ and $g_2$ generate $G$. Show that #fixed $g_1$ = # fixed $g_2$.

I first showed that the number of generators of $Z_n$ was $\phi(n)$ but I don't know how to proceed. Any help would be great, thank you in advance!

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  • $\begingroup$ Well, first show that $\langle k\rangle$ has order $d$, then show that any subgroup of order $d$ must contain $k$. $\endgroup$ – Don Thousand Mar 26 '19 at 21:14
  • $\begingroup$ Yes, @DonThousand, that was my thought too but I'm not sure how to go about either of those steps. $\endgroup$ – James Done Mar 26 '19 at 21:17
  • $\begingroup$ The first part is pretty easy ... Think about multiples of $k$. The second part, note that subgroups of cyclic groups are cyclic... $\endgroup$ – Don Thousand Mar 26 '19 at 21:19
  • $\begingroup$ Ok, I think I got the first part now. But I still don't understand how to do the "must contain k" part of it. Sorry, thank you for your time! $\endgroup$ – James Done Mar 26 '19 at 21:23
  • $\begingroup$ Note that any subgroup of order $d$ must contain an element of order $d$. As such, it must be $g^k$ for some generator $g$. What should that tell you? $\endgroup$ – Don Thousand Mar 26 '19 at 21:24
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I first showed that the number of generators of $\mathbb{Z}/n$ was $\phi(n)$ ...

This has essentially no relevance to the problem. Counting generators won't help, because that's not what we're interested in.

For the first part: since the multiples of $k$ are a subgroup of $\mathbb{Z}$, their reductions mod $n$ are a subgroup of the integers mod $n$. How many different multiples of $k$ are there mod $n$?
Uniqueness is an odd thing to be asking about here - we've specified one particular subgroup already, in a way that didn't leave any choices. Without more context, I'm not sure what "a unique subgroup" is supposed to mean here.

For the second part: since $g_1$ generates the group, $g_2=g_1^m$ is some power of $g_1$. Now, if $x$ is a fixed point for the action of $g_1$, show that it's also a fixed point for the action of $g_2$.
Then, switch. Since $g_2$ generates the group...

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  • $\begingroup$ Ok, this makes sense but how would I show that if x is a fixed point for the action of $g_1$, it is also a fixed point action for the action of $g_2$ (my textbook doesn't explain fixed points at all). Thank you @jmerry! $\endgroup$ – James Done Mar 26 '19 at 21:54

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