So I want to show that $\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle$. My approach to this problem was to show a double containment, i.e. to show that $\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle$ and $ \langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle$.
I would like to see a full proof for this, specifically $\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle$. (I tried it with the approach of breaking it down into to cases; $a$ has infinite order and $a$ has finite order, the latter of which i would appreciate the most help on.)
My approach to solving the whole problem: (I would appreciate any feedback on anything that is wrong, or a different approach to the proof.)
To show the easier containment, $\langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle$, I did the following: Let $l = \operatorname{lcm}(m, n)$. Let $j \in \langle a^l\rangle$, so $j = (a^l)^k = a^{lk}$ for some $k \in \mathbb Z$. Since $l$ is a multiple of $m$ and $n$ by definition, we can say $l = ms = nt$ for some $s, t \in \mathbb Z$. Now $j = a^{kl} = a^{kms} = (a^m)^{ks} \in \langle a^m\rangle$. Similarly, $j = a^{kl} = a^{knt} = (a^n)^{kt} \in \langle a^n\rangle$. Now, since $j \in \langle a^m\rangle$ and $j \in \langle a^n\rangle$, it follows that $j \in \langle a^m \rangle \cap \langle a^n \rangle$. Thus, by definition, $\langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle$.
For the second containment, the one which I'm having more problems with, I attempted the following:
Case in which $a$ is infinite: Suppose that $\vert a \vert = \infty$. Let $c \in \langle a^m \rangle \cap \langle a^n \rangle$. Then $c = a^{mx} = a^{ny}$ where $x, y \in \mathbb Z$. It follows that $a^{mx - ny} = e$ and so $mx = ny$ because if $mx > ny$ then the difference would not be zero, and we would have an element that was finite, contradicting our hypothesis. And since $mx = ny$ we know $\operatorname{lcm}(mx, ny) = mx = ny$ and $\operatorname{lcm}(mx, ny)$ is a multiple of $\operatorname{lcm}(m, n)$. Hence $c \in \langle a^{\operatorname{lcm}(m, n)}\rangle$ and thus $\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle$.
Case in which $a$ is finite: I tried starting it out the same as the previous case, but i could never reach my conclusion :(
$\def\lcm{\mathrm{lcm}}$Let $k$ be the order of $a$, with $k=0$ if $a$ is of infinite order. Note that $x\equiv y \pmod{0}$ is equivalent to $x=y$ (since $0|x-y$ if and only if $x-y=0$). Also, $\mathrm{lcm}(n,0) = \gcd(n,0) = n$.
Consider first the case with $n|k$ and $m|k$.
To show $\langle a^m\rangle\cap\langle a^n\rangle \subseteq \langle a^{\lcm(m,n)}\rangle$, let $x$ lie in the intersection. Then there exist $r,s\in\mathbb{Z}$ such that $x=a^{mr} = a^{ns}$, so $mr\equiv ns\pmod{k}$. Thus, $k|mr-ns$, so $n|mr-ns$ and $m|mr-ns$. Therefore, $n|mr$ and $m|ns$. Since $n|mr$, then $mr$ is a common multiple of $n$ and $m$, hence is a multiple of $\lcm(m,n)$ (similarly with $ns$), so we can find $q$ such that $mr=q\lcm(m,n)$. Thus, $x = a^{mr} = a^{q\lcm(m,n)}\in\langle a^{\lcm(m,n)}\rangle$, as desired.
Now for the general case, note that $\langle a^n\rangle = \langle a^{\gcd(n,k)}\rangle$; thus, we are reduced to showing that $\lcm(n,m) \equiv \lcm(\gcd(n,k),\gcd(m,k))\pmod{k}$. This is true, but you may not have it in your arsenal yet, so here's a proof (that no doubt Bill Dubuque can prove in a much slicker way):
Proof. Indeed, $\gcd(n,k)$ and $\gcd(m,k)$ both divide $k$, hence their least common multiple divides $k$; and $\lcm(n,m)$ is a common multiple of $\gcd(n,k)$ and $\gcd(m,k)$, so $\lcm(\gcd(n,k),\gcd(m,k))$ divides $\lcm(n,m)$. Thus, the left hand side divides the right hand side. Conversely, let $p$ be a prime and suppose that $p^a$ is the largest power of $p$ that divides $\gcd(\lcm(n,m),k)$. Then $p^a$ divides both $\lcm(n,m)$ and $k$, but $p^{a+1}$ does not divide at least one of them. Since $p^a$ divides $\lcm(n,m)$, then $p^a$ divides either $n$ or $m$, and we also know $p^a$ divides $k$. Either $p^{a+1}$ does not divide $k$, or it does not divide either $n$ or $m$. Thus, $p^a$ divides either $\gcd(n,k)$ or $\gcd(m,k)$, but $p^{a+1}$ divides neither. Therefore, $p^a$ is the largest power of $p$ that divides $\lcm(\gcd(n,k),\gcd(m,k))$. $\Box$
Now the result follows: note that $\gcd(\lcm(n,m),k) \equiv \lcm(n,m)\pmod{k}$ (express $\gcd(\lcm(m,n),k)$ as a linear combination of $\lcm(m,,n)$ and $k$), so we are done.