Prove that $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{lcm (n,k)} \rangle$

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?

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.

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