Let $G=<a>$ be a cyclic group. $O(a)=24$. We will define $H_k$: $H_k=<a^k>$. Find for what $k$ values, $0\leq k \leq 23$, $a^5H_k=a^{13}H_k$.
I couldn't figure a way to answer this quickly. I know that $a^5H_k=a^{13}H_k \iff a^5\in a^{13}H_k$.
Obviously it will be equal for all $k$ such that $gcd(k,24)=1$. It will also be equal if $k|16$. Beyond that I am unable to tell (without checking all values one by one...). Is there an efficient way to solve this problem?
On
As you've noted, your question is equivalent to finding solutions to
$$nk \equiv 16\pmod{16}$$
since $a^5\in a^{13}H_k$ means $a^{8+nk} = a^0$. In general, an equation
$$a_1x_1 + \dots a_nx_n\equiv b\pmod{m}$$
has solutions if and only if $\gcd(a_1,\dots, a_n, m)~|~b$. Thus, the equation $nk\equiv 16\pmod{24}$ has solutions for $n$ when $\gcd(k,24)~|~16$. Since $24 = 2^3\cdot3$, the only obstruction to $k$ being a viable option is if $k\equiv 0\pmod{3}$. Thus, the only $k$ that do not work are
$$3,6,9,12,15,18,21.$$
As a general statement, let $G=\langle a\rangle$ where $|a| = m$. If we want to find the $k$ such that $a^sH_k = a^tH_k$, we're really solving the equation
$$nk \equiv s-t\pmod{m}$$
which has a solution for $n$ when $\gcd(k,m)~|~(s-t)$.
Hint: $a^5H_k = a^{13}H_k$ if and only if $a^8\in H_k = \langle a^k\rangle$.