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The title says it all. I am trying to solve a homework problem, the ultimate goal of which is to prove that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic for prime $p$. (This is from Herstein's book, if you are interested). I already solved the previous parts, one of which is if $G$ is a finite abelian group such that $x^m=e$ has at most $m$ solutions for any $m$ dividing $n$, then $G$ is cyclic. I am trying to apply this to $(\mathbb{Z}/p\mathbb{Z})^{\times}$ with order $p-1$. This would require showing that $x^m=1$ has at most $m$ solutions, which would prove that $(\mathbb{Z}/p\mathbb{Z})^{\times}$. One thing I noticed is the set $\{x\in (\mathbb{Z}/p\mathbb{Z})^{\times}: x^m=e\}$ is a subgroup of $(\mathbb{Z}/p\mathbb{Z})^{\times}$. I was thinking that I can show that this subgroup cannot have more than $m$ elements using Lagrange's Theorem. The problem is, I do not know anything about the divisors of $p-1$, nor do I ever use the fact that $p$ is prime. I cannot use anything other than Lagrange's Theorem and basic facts about groups and subgroups, as well as Euler's Theorem and Fermat's Little Theorem.

Any suggestions? Please do not give full solutions, but hints are welcome.

Shaun
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Aadi Rane
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  • You can use the following fact for every positive integer $n$, that $n=\sum_{d|n} \varphi(d)$, for all divisors $d$ of $n$, where $\varphi$ is the Euler's function. – wrath Feb 08 '24 at 06:27
  • How does that help me show that $x^m=1$ has at most $m$ solutions? – Aadi Rane Feb 08 '24 at 06:47
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    At some point you have to use the properties of the field $\mathbb{Z}/p\mathbb{Z}$, because there may be other abelian groups of the same size $p-1$ and in those other groups it may not be the case that $x^m=e$ has only $m$ solutions, so in order to get the result you have to rely on specific features of the structure of $\mathbb{Z}/p\mathbb{Z}$ – Chris Sanders Feb 08 '24 at 08:37
  • Maybe a matter of taste, but for me the way to prove this is to establish that $\mathbb{Z} / p\mathbb{Z}$ is a field and then argue from the fact that a degree $m$ polynomial over a field has at most $m$ roots. Each fact can be pretty quickly and straightforwardly proved. – Charles Hudgins Feb 09 '24 at 02:25
  • @AadiRane does my answer satisfy your question? – Krave37 Feb 12 '24 at 09:45

1 Answers1

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Edit: I do not have much experience with abstract algebra so if I used something that you had mentioned not to please comment.

Hint-1

Think about the number of $x \pmod{p}$ having some order $d$ where $d$ divides $p-1$.This is well known.

Hint-2

If there exists an $ord_p(x) = z$ where $z|m$, then $x^m \equiv 1 \pmod{p}$

Proceed iff hints fail

Solution :

Answer to hint one is $\phi(d)$, the proof can be found here -

Prove that exactly $\phi(d)$ elements have order $d$

When $d \mid n$, the number of elements of order $d$ in a cyclic subgroup of order $n$ is $\phi (d)$

The total solutions to $x^m \equiv 1 \pmod{p}$ will be the sum of all $\phi(d)$ where $d|m$ and $d|(p-1)$ (why?,comment if you have a problem with this statement).

This means that $\sum_{d|m}\phi(d)= m$(This is also well known and is $m$ itself) is an upper bound, as it may or may not satisfy $d|p-1$.

It will satisfy when $m|p-1$.

Therefore, Number of solutions $\leq m$

Krave37
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  • The proof in your link to showing that exactly $\phi(d)$ elements have order $d$ assumes that $({\mathbb Z}/p{\mathbb Z})^\times$ is cyclic, which we are not allowed to use (that is what the OP is trying to prove). – Derek Holt Feb 08 '24 at 21:50