Order Free Proof of $\langle a^m\rangle\cap \langle a^n \rangle=\langle a^{\text{lcm}(n,m)}\rangle$ There are clear proofs of the result that $\langle a^m\rangle\cap \langle a^n \rangle=\langle a^{\text{lcm}(n,m)}\rangle$ here, but I'm wondering if the following proof is correct since it avoids any reference to the order of $\langle a \rangle$ or the notion of $\text{ lcm}(\gcd(k,m),\gcd(k,n))$.
Proof:
Let $s=\text{lcm}(m,n)$. Then for some $t_0,t_1\in\mathbb{Z}$ we have $s=mt_0$ and $s=nt_1$. Thus if $g\in\langle a^s\rangle$, then $g=(a^s)^t$ for some $t\in\mathbb{Z}$ so that $$g=(a^s)^t=(a^{mt_0})^t=(a^m)^{tt_0}\in\langle a^m\rangle$$
and $$g=(a^s)^t=(a^n)^{tt_1}\in\langle a^n\rangle$$
therefore $g\in\langle a^m\rangle\cap \langle a^n \rangle$.
On the other hand, if $g=a^r\in\langle a^m\rangle\cap \langle a^n \rangle$, then $m$ and $n$ divide $r$ so that $\text{lcm}(m,n)$ divides $r\implies g=a^r\in\langle a^s\rangle$.
Hence $\langle a^m\rangle\cap \langle a^n \rangle=\langle a^s \rangle =\langle a^{\text{lcm}(n,m)}\rangle$. $\blacksquare$
I'm a little concerned because this proof does not seem to invoke the order of $\langle a \rangle$, and when I compute these groups assuming finiteness I first reduce the powers $a^m$ and $a^n$ via the G.C.D. of the powers of $a$ and the order of $a$ so that $\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\gcd(|a|,m)} \rangle \cap \langle a^{\gcd(|a|,n)} \rangle$, and then compute $$\langle a^{\gcd(|a|,m)} \rangle \cap \langle a^{\gcd(|a|,n)} \rangle=\langle a^{\text{lcm}(\gcd(|a|,m),\gcd(|a|,n))\text{mod}(|a|)}\rangle $$
so that we have $$\langle a^m \rangle \cap \langle a^n \rangle=\langle a^{\text{lcm}(\gcd(|a|,m),\gcd(|a|,n))\text{mod}(|a|)}\rangle.$$
How can this all be suppressed in the proof I gave above?
 A: The argument presented is incorrect/incomplete.
The argument that $a^{\mathrm{lcm}(m,n)}\in \langle a^n\rangle \cap \langle a^m\rangle$ is fine.
But the converse inclusion has a gap. You say:

On the other hand, if $g=a^r\in\langle a^m\rangle\cap\langle a^n\rangle$, then $m$ and $n$ divide $r$.

This conclusion does not follow without additional hypotheses (e.g., that $a^r=a^s\iff r=s$). For example, consider the case where $a$ has order $10$, $m=8$, and $n=6$. Then
$$\begin{align*}
\langle a^8\rangle &= \{ a^8, a^6, a^4, a^2, 1\},\\
\langle a^6\rangle &= \{ a^6, a^2, a^8, a^4, 1\}.
\end{align*}$$
Therefore, $a^4\in\langle a^8\rangle\cap \langle a^6\rangle$, but it is false that $8$ and $6$ both divide $4$. What is true is that there exists an integer $k$ such that $4\equiv k\pmod{10}$ with $8$ and $6$ both dividing $k$ (namely, $24$ in this case). Likewise with each of the elements $a^2$, $a^6$, and $a^8$, all of which lie in the intersection, but none of which have exponents that are multiples of both $6$ and $8$.
So that is the gap where you are missing the consideration of the order of $a$. The argument you give does not work as is.
