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I am hoping for verification of this proof. My concern is if the contradiction really refutes that k does not divide n.

PROOF

Let $a\in G$ where $G$ is a group. Let $k$ be the smallest positive integer such that $a^k=1$. Let $n\in\mathbb{Z^+}$, such that $a^n=1$.

This implies that $k\le n$. If $k=n$ and $p=1$, then $n=kp$. Thus $k$ divides $n$.

Now consider $k<n$.

If $k=1$, then $a^k=a^1=1$. Choose $m=n$. Then $n=m=1m=km$ and $k$ divides $n$.

Now assume $k>1$ and assume $\forall q\in\mathbb{Z}$, $n\ne kq$. Since $1<k<n$ we can exclude the cases on $q\le 0$ and we consider only $q>0$.

Choose a $q$ such that $kq<n$ and $k(q+1)>n$. This $q$ exists in view of the division algorithm, $n=kq+r$, where $r\in\mathbb{Z}$ and $0\le r < \vert k\vert = k$. This implies that $kq=n-r<n$ and $n< kq+k = k(q+1)$.

Subtracting $kq$ from both sides of the latter inequality gives $0<n-kq<k$.

Thus, we have $a^n=1=1^q=(a^k)^q=a^{kq}$. This implies that $a^{n-qk}=1$. Since $n-kq<k$, this implies that k is not the smallest positive integer such that $a^k=1$, a contradiction.

Hence, there exists a $q\in\mathbb{Z}$ such that $n=kq$. Therefore $k$ divides $n$.

RJM
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    Can't you just write $n=kq + r$ where $0\leq r<k$ and then $a^n = (a^k)^q*a^r = a^r = 1$ and since $r < k$, $r=0$? – Snowball May 06 '21 at 17:27
  • If you know about subgroups of $\mathbb Z$, then the set ${ n \in \mathbb Z : a^n =1}$ is a nontrivial subgroup of $\mathbb Z$ and so is generated by its smallest positive element. – lhf May 06 '21 at 17:35
  • @Snowball yes. But I should have had $0<r<k$. Thank you. – RJM May 06 '21 at 18:44
  • @Ihf not that far along yet. Just started with group theory. Thank you, though. – RJM May 06 '21 at 18:44
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    The proofs in the linked dupes (by Euclidean descent) also work here. You are on the way to rediscover them (in less concise form). – Bill Dubuque May 07 '21 at 00:58
  • @Bill Dubuque thank you. That is excellent. I actually wanted to prove this statement for a problem that asked whether or not it $k$ divides $n$, but did not ask for the proof. Basically, in chapter 1.1 of the abstract algebra book I just started. I am just happy I was on the right track, which it seems was so, despite an unoptimized proof. Your link, however, is very informative. – RJM May 07 '21 at 01:37
  • Glad it helped. I expanded this answer. – Bill Dubuque May 07 '21 at 01:53
  • @Bill Dubuque so cool. I am just learning real mathematics myself (I am a chemist and pharmacist). Such examples are why I find it so amazing. Thanks again. – RJM May 07 '21 at 01:55

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You can simplify the proof. If $k \nmid n$, then by the division algorithm $\exists q, r \in \Bbb Z^+~(n=kq+r)$, where $0 \lt r \lt k$. But then $1=a^n= a^{kq} a^r=({a^k})^q a^r=1^qa^r=a^r $, contradicting the minimality of $k$.

Robert Shore
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  • Thank you. And, in turn, showing that contradiction provides for divisibility, as non-divisibility was implied in the $r>0$. – RJM May 06 '21 at 18:38