Show $\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle$ 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 :(
 A: Note $ $ The equality in the lemma in Arturo's answer - that gcd distributes over lcm - can be derived simply via lcm, gcd laws. Rewriting lcms to gcds using $\rm\,[a,b] := lcm(a,b) =\: ab/(a,b)\,$ and clearing denominators, we deduce $\,\rm (k,[n,m]) = [(k,n),(k,m)]\,$ is equivalent to
$$\rm (k,m,n)\:(km,kn,mn)\ =\ (k,m)\:(k,n)\:(m,n)$$
whic is true, by both $\rm = (kkm,kkn,kmn,kmm,knn,mmn,mnn)\:$ by the distributive law. $ $ QED
For further details on the above gcd arithmetic this answer.
Dually, we could rewrite gcds to lcms, reducing it to the lcm dual of the above. And, of course, the dual identity is true: lcm distributes over gcd. These are well-known identities that come to the fore in the lattice viewpoint, e.g. see Birkhoff, Lattice Theory.
Though it is no longer needed after the above, it is instructive to note that Arturo's proof of lhs|rhs follows nicely from a useful lemma. First, recall the  universal properties of lcm, gcd
$$\rm\
 [a,b]\:|\:n \;\iff\; a,b\:|\:n\quad where\quad\: [a,b] := lcm(a,b) $$
$$\rm n\:|\:(c,d) \;\iff\; n\:|\:c,d\quad where\quad (c,d) := gcd(c,d) $$
Employing these we immediately deduce the following "lcm divides gcd" lemma
$$\rm [a,b]\ |\ (c,d)\iff a,b\ |\ (c,d)\iff a,b\ |\ c,d $$
Now, applying this to the particular values in Arturo's lemma we deduce
$$\rm [(n,k),(m,k)]\ |\ (k,[n,m])\iff (n,k),(m,k)\ |\ k,[n,m]$$
which is true: $\rm\:(n,k),(m,k)\ |\ k\:$ and $\rm\:(n,k)\ |\ n\ |\ [n,m],\:$ and $\rm\: (m,k)\ |\ m\ |\ [n,m]\quad$ QED
A: $\def\lcm{\mathrm{lcm}}$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):

Lemma. $\lcm(\gcd(n,k),\gcd(m,k)) = \gcd(\lcm(n,m),k)$. 

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.
