How to show $\langle a^k\rangle\leq \langle a^\ell\rangle\Leftrightarrow \gcd(\ell, n)\mid \gcd(k, n).$? Consider the cyclic group $G=\langle a\rangle$ where $o(a)=n$ where $o(a)$ means order of $a$. I'd like to show: $$\langle a^k\rangle\leq \langle a^\ell\rangle\Leftrightarrow \gcd(\ell, n)\mid \gcd(k, n).$$
I have already done the implication $(\Rightarrow)$ as follows:
I know $$\displaystyle o(a^k)=\frac{n}{\gcd(k, n)}\quad \textrm{and}\quad \displaystyle o(a^\ell)=\frac{n}{\gcd(\ell, n)}.$$ By Lagrange's theorem  if $\langle a^k\rangle\leq \langle a^\ell\rangle$ then $o(a^k)\mid o(a^\ell)$ hence $\gcd(\ell, n)\mid \gcd(k, n)$.
How can I show the opposite implication?
Thanks

Let $\gcd(l,n)=d_1$ and $\gcd(k,n)=d_2$, now $d_1=a_1l+b_1n$ and $d_2=a_2k+b_2n$  for some $a_1,b_1,a_2,b_2 \in \mathbb{Z}$ and we have $d_1|d_2$ so it implies $a_1l+b_1n=t(a_2k+b_2n)$  for some $t \in \mathbb{Z}$  hence $$ta_2k=(a_1l+b_1n-tb_2n)$$ so, $$a^{ta_2k}=a^{(a_1l+b_1n-tb_2n)}=a^{a_1l} \in <a^l>$$ and so $$a^{k-ta_2k}.a^{ta_2k}=a^k \in <a^l>$$ $\hspace{16cm}$$\mathbf{QED}$
