TITLE: Show $\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle$
QUESTION [18 upvotes]: 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 :(

REPLY [3 votes]: $\newcommand{\lcm}{\operatorname{lcm}}$
Lemma. Let $G=\langle g\rangle$ a cyclic group. If $H$ is a subgroup of $G$, then $H$ is cyclic. Moreover if $H\neq \{e\}$ then $$H=\left\langle g^{\min\{j\in\mathbb{N}:g^j\in H\}}\right\rangle.$$
Proof. If $H=\{e\}$ ($e$ for the identity element) then $H$ is cyclic. Suppose that $H\neq \{e\}$. Then $H\subseteq \{g^n:n\in\mathbb{Z}\}$, i.e. there exist $n\in\mathbb{Z}^*$ so that $g^n\in H$. Note that $n\in\mathbb N$ or $-n\in\mathbb N$, in any case $\{j\in\mathbb{N}:g^j\in H\}\neq\emptyset$, so $$k=\min \{j\in\mathbb{N}:g^j\in H\}$$ is well defined. Now clearly, $\langle g^k\rangle\subseteq H$. In the other hand, if $g^n\in H$ for some $n\in\mathbb{Z}$ since $k\neq 0$ there exist $q,r\in \mathbb{Z}$ so that $$n=qk+r,\qquad 0\leq r\lt k.$$ If $1\leq r\lt k$ then
$$\begin{align*}
g^n &= g^{qk}g^r\\
g^ng^{-qk}&= g^r,
\end{align*}$$
since $g^{qk}=(g^k)^q\in H$, this says that $g^r\in H$ which contradicts the election of $k$. Therefore $$g^n=g^{qk}=(g^k)^q\in\langle g^k\rangle$$
as we wanted. Therefore if $H\neq \{e\}$, $H=\langle g^k\rangle.$
In your case, you have $$ a^{\lcm(m, n)} \in \langle a^m\rangle\cap\langle a^n\rangle.$$ Now, note that $a^j\in \langle a^m\rangle\cap\langle a^n\rangle$ iff $a^j\in \langle a^m\rangle$ and $a^j\in \langle a^n\rangle$ iff $m|j$ and $n|j$. Thus $$\{j\in\mathbb{N}:a^j\in \langle a^m\rangle\cap\langle a^n\rangle\}=\{j\in\mathbb{N}:m|j\text{ and } n|j\},$$ and therefore $$\min\{j\in\mathbb{N}:a^j\in \langle a^m\rangle\cap\langle a^n\rangle\}=\lcm(m,n)$$ by the definition of $\lcm$. Therefore $$\langle a^m\rangle\cap\langle a^n\rangle=\langle a^{\lcm(m,n)}\rangle.$$