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Let $\color{brown}{C}$ be a finite cyclic group, and $\color{brown}{a\in C}$ , such that $\color{brown}{|a|=n}$, then $$\color{green}{|a^k|=\frac{n}{gcd(k,n)}=\frac{lcm(k,n)}{k}}$$How to intuitively think/justify the above mathematical statement. $$\color{grey}{any\quad help\quad is \quad deeply\quad appreciated}$$

user766817
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  • Just play around with the case $n=6$, say, to get a feeling? – Hagen von Eitzen May 20 '20 at 05:38
  • Can you intuitively understand it in the special case of a permutation? If so, then the theorem follows from Cayley's theorem (every group is a group of permutations) – HallaSurvivor May 20 '20 at 06:09
  • Looking for something where we talk about role of gcd/LCM in deciding the least positive power for a^k – user766817 May 20 '20 at 06:17
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    I think, a decent proof helps a lot for "intuitively think/justify". See this duplicate. Then of course some examples with small $n$ are helpful. – Dietrich Burde May 20 '20 at 08:13
  • The closest thing I could come with is you should convince yourself that $\mid a^k \mid = n$ for $n,k$ relatively prime. And because of the minimality of the order, for the general case it is necessary to divide $n$ by the gcd. The second and third equality is pretty trivial. – Minh Khôi May 20 '20 at 11:03
  • If $\mid a^k \mid = h < n$ then you could take $hk$ by the modulo $n$ and end up with an order for $a$ that is less than $n,$ that is a contradiction. You could also verify that $a^{k^n} = e.$ – Minh Khôi May 20 '20 at 11:10
  • All the comments are helpful. The last comment which states result $a^{k^n}=e$ looks dubious as that $\Rightarrow n|{k^n} $ which is not always possible. Please correct me if I am wrong. – user766817 May 21 '20 at 23:34
  • The possibility holds true $\iff lcm(k^n,n}=k^n$ – user766817 May 21 '20 at 23:41

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