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I was reading the theorem about the existence of an integer $t$, the primitive root modulo prime. The proof seemed a bit confusing. I mean the construction part. Why did not they immediately take $t = xy$ instead of $t = x^{m'}y^{m}$? I think $xy$ also satisfies the requirements. Thanks in advance. Here is the link of the proof:

http://www.math.stonybrook.edu/~scott/blair/Proof_Theorem_5.html#B

shota kobakhidze
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    I think you can write down a quick sketch here and underline the problem that you want to ask to be more precise. It would help you to get an answer quickly. :) – Kumar Jul 08 '19 at 14:18
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    That's very difficult to read. I suspect though that $xy$ need not necessarily work. – Angina Seng Jul 08 '19 at 14:32
  • I will make things easier here. So I have an integer $x$ and $d$ is the smallest integer such that $x^{d} \equiv 1$ (mod p). We have another integer $y$ and the smallest integer $e$ such that $y^{e} \equiv 1$ (mod p). Then We want to construct an integer $t$ for which$f = LCM(d, e)$ is the smallest integer such that $t^{f} \equiv 1$ (mod p). That is what I got from there. So what if we take $t = xy$? $x^{k} \equiv 1$ (mod p) if and only if $k$ is a multiple of $d$ and $y^{l} \equiv 1$ (mod p) if and only if $l$ is divisible by $e$. So $LCM(d, e)$ really satisfies the requirements. – shota kobakhidze Jul 08 '19 at 14:47

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Generally it is not true that in an abelian group that if $\,x,y\,$ have order $\,j,k\,$ then $xy$ has order $\,{\rm lcm}(j,k),\,$ e.g. consider the case $\,y = x^{-1}.\,$ But it is true that there exists some element of order $\,{\rm lcm}(j,k),\,$ and this is what is proved there (see here for a few other proofs of order lcm-closure)

Remark $ $ Their proof can be simplified. By here: $ $ if $\,x,y\,$ have order $\,d,e\,$ then there are coprime $\,m',m\in \Bbb N\,$ with $\,(d,e)={m'}\,{m},\ (d/m',\,e/m)=1\,$ so $\,x^{\large m'},\, y^{\large m}$ have coprime orders $\,d/m',\, e/m\,$ therefore their product has order $\ (d/m')(e/m) = de/(d,e) = {\rm lcm}(d,e)$.

Unlike many proofs, the linked proof does not require expensive prime factorization. Instead it employs only gcds so it yields an efficient algorithm to compute $\,m',m.$

Bill Dubuque
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  • Note: I checked the outline of the cited proof but I did not verify its correctness. – Bill Dubuque Jul 08 '19 at 18:52
  • Thanks for the response. So, as I understood their proof is complicated because of the formal correctness reasons from the point of view of group theory? However, in this case $xy$ works but such reasoning could fail in some similar situations? – shota kobakhidze Jul 08 '19 at 20:43
  • @shotakobakhidze Why do you believe that $xy$ works? – Bill Dubuque Jul 08 '19 at 20:59
  • Yes, you are right. I realized that it is incorrect if in some step I have only left $y$ which is inverse of $x$ and that fails as you mentioned. Then the answer for them is not $LCM(d, e)$ but 1. Nice point. – shota kobakhidze Jul 08 '19 at 21:13
  • @shota kobakhidze: The author avoids the issue of $y$ being the inverse of $x$ by choosing $y$ so that the order of $y$ doesn't divide $d$. Since the equation $x^d=1$ has at most $d$ solutions, such a choice is always possible (assuming $d < p-1$, we have $d\le{\large{\frac{p-1}{2}}}$). – quasi Jul 08 '19 at 21:28
  • Yes, even it is not necessary $y$ to be an inverse of $x$ to fail my assumption. Even if $x^{k}$ is inverse of $y^{k}$ , $k < LCM(d, e)$, my assumption easily fails in this case too. – shota kobakhidze Jul 08 '19 at 21:37
  • @quasi Yes, my point was merely to give a quick simple example where the general claim fails in order to nudge the OP to debug their argument. It is easy to exhibit other counterexamples once one knows that it fails in general. – Bill Dubuque Jul 08 '19 at 21:39
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Let $x\in \{1,...,p-1\}$, and let $d$ be the order of $x$.

If $d=p-1$, then $x$ is a primitive root, and we're done.

Suppose $d < p-1$.

The plan is to find some element of $t\in\{1,...,p-1\}$ whose order exceeds $d$, and then iterate, using $t$ as the new $x$.

As the author argues, there exists $y\in\{1,...,p-1\}$ whose order doesn't divide $d$.

Let $e$ be the order of $y$.

If $e > d$, we can let $t=y$.

Since $e\not\mid d$, we can't have $e=d$.

Suppose $e < d$.

Your claim is that we can let $t=xy$.

Unfortunately, this doesn't always work.

As you correctly observed, since $e\not\mid d$, we get $\text{gcd}(d,e) < e$, hence $$ \text{lcm}(d,e) = \frac{de}{\text{gcd}(d,e)} = d\left(\frac{e}{\text{gcd}(d,e)}\right) > d $$ Let $f$ be the order of $xy$.

Clearly $f{\,|\,}\bigl(\text{lcm}(d,e)\bigr)$.

However, noting Bill Dubuque's post, and correcting my earlier answer, it's not automatic that $f=\text{lcm}(d,e)$.

In fact, we can't even claim $f > d$.

As an example, letting $p=31,x=7,y=23$, we get

  • $x$ has order $d=15$.$\\[4pt]$
  • $y$ has order $e=10$.$\\[4pt]$
  • $xy$ has order $f=6$.

This shows that your idea of using $xy$ for the next iteration doesn't always work.

quasi
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