Prove that $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{lcm (n,k)} \rangle$ Let $G$ be a group. Let $a$ be an element. Let $n,k$ be pozitive integers. Let $m$ be least common multiple of $n$ and $k$. Prove $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{m} \rangle$. Trying to show both are subsets of each other. One side is trivial. But I can not show that $\langle a^n \rangle$ intersection $\langle a^k \rangle$ is a subset of $\langle a^m \rangle$
 A: Consider the canonical surjective homomorphism $\pi:\mathbb Z \to A=\langle a \rangle$ given by $t \mapsto a^t$.
Let $A_t = \langle a^t \rangle$ for $t \in \mathbb N$. All subgroups of $A$ are of this form. The lattice of subgroups of $A$ is isomorphic to the lattice of subgroups of $\mathbb Z$ containing $\ker \pi$.
If the order of $a$ is infinite, then $\pi$ is an isomorphism taking $t\mathbb Z$ to $A_t$. The $A_t$ are all different and the result follows from the fact that the lattice of subgroups of $\mathbb Z$ reflects divisibility. In particular, $n \mathbb Z \cap k \mathbb Z = lcm(n,k) \mathbb Z$.
If the order of $a$ is finite, the subgroups $A_t$ cannot be all different, since there are only a finite number of them. They are all different only when we take $t$ to be a divisor of $N$, the order of $a$. For such $t$, the lattice of subgroups of $A$ is isomorphic to the lattice of subgroups of $\mathbb Z$ that contain $N\mathbb Z$, which reflects divisibility among the divisors of $N$.
In general, $A_t = A_{gcd(N,t)}$. This is the key point. It follows from writing $gcd(N,t)=uN+vt$.
The result then follows from the distributivity property of gcd and lcm:
$$
gcd(u, lcm(v, w)) = lcm(gcd(u, v), gcd(u, w))
$$
Indeed,
$$
A_n \cap A_k = A_{gcd(N,n)} \cap A_{gcd(N,k)} = A_{lcm(gcd(N,n),gcd(N,k))} = A_{gcd(N,lcm(n,k))} = A_{lcm(n,k)}
$$
A: Let $g$ be in the intersection, i.e. $g=(a^n)^r=a^{nr}$ and $g=(a^k)^s=a^{ks}$. This means $a^{nr}=a^{ks}$. Can you take over from here?
A: You took $lcm (n,k)=m$
Now, let $x=a^t \in \langle a^n \rangle \bigcap \langle a^k \rangle$
Then, $n$ divides $t$ and $k$ divides $t$. 
So, $m=lcm(n,k)$ divides $t$
i.e. $x=a^t\in \langle a^m \rangle$
For opposite containment
Let $x=a^t\in \langle a^m \rangle$
Then, $m$ divides $t$
Now, since $n$ and $k$ divide $m$ you have the containment.
So equality holds.
