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 :(
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keys. Further, this is not a problem that specialists' skills are required to track. This is a typical problem in a first course in group theory. Regards, – Apr 15 '12 at 21:12