Let $a^n, a^m \in (a^k)$ for some positive integer $k$. Then $k \mid n, m.$ Hence $k \mid \operatorname{gcd(n, m)}?$

Question

Let $a^n, a^m \in (a^k)$ for some positive integer $k$. Then $k \mid n, m.$ Hence $k \mid \operatorname{gcd(n, m)}$.

How is it possible? Let $k = 6, n = 12, m =24.$

Context:

Let $H$ be the smallest subgroup of $(a)$ s.t. $a^n, a^m \in H$. Prove that $H = (a^{\operatorname{gcd(n, m)}}).$

Proof: Since $G$ is cyclic, $H$ is cyclic. Hence $H = (a^k)$ for some $k \in \mathbb N.$ Since $a^n, a^m \in H,$ then $k \mid n,m.$ Hence $k \mid \operatorname{gcd(n, m)}.$ Thus $a^{\operatorname{gcd(n, m)}} \in H.$ Hence $(a^{\operatorname{gcd(n, m)}}) \subset H.$ Also, since $a^{\operatorname{gcd(n, m)}} \mid n, m$ then $a^n, a^m \in (a^{\operatorname{gcd(n, m)}})$. Since $H$ is the smallest subgroup of $(a)$ containing $a^n, a^m$ and $a^n, a^m \in (a^{\operatorname{gcd(n, m)}})$, we conclude that $H = (a^{\operatorname{gcd(n, m)}}).$

In a group, this is only technically true if the order of $a$ is divisible by $k$, since otherwise, it is possible for $a^n\in(a^k)$ when $n$ is not divisible by $k$. — Thomas Andrews, Apr 21 '16 at 19:46
A cyclic group of finite order, or a cyclic group of infinite order? — Thomas Andrews, Apr 21 '16 at 21:52
@ThomasAndrews, that I don't know. My book says nothing about that. — user333573, Apr 21 '16 at 22:07
The followup context actually doesn't matter if the group is finite or infinite, but your statement is not true for finite groups unless the order of $a$ is divisible by $k$, as I mentioned before. — Thomas Andrews, Apr 21 '16 at 22:23

score 1 · Accepted Answer · answered Apr 21 '16 at 19:33

Since $k$ divides both $m$ and $n$, hence it divides all linear combinations $mx+ny$. GCD is the least positive linear combination of $m$ and $n$, hence divisible by $k$ as well OR you can use the definition of GCD which stipulates that every common divisor (which happens to be $k$ here) divides the GCD.

Let $a^n, a^m \in (a^k)$ for some positive integer $k$. Then $k \mid n, m.$ Hence $k \mid \operatorname{gcd(n, m)}?$

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