We show that (a) implies (b). Since $k$ and $n$ are relatively prime, $kp + nq = 1$ for some integers $p$ and $q$. Let $H$ be a subgroup of $G$, and let $x$ be a member of $G$ such that $x^k \in H$. Since $x^n = e$, the unit of $G$, it follows that
$$x = x^{kp + nq} = (x^k)^p \cdot (x^n)^q = (x^k)^p \in H.$$

We now show that (b) implies (a). This is clearly the case if $k = 1$ or $n = 1$, so let them both be at least $2$, and suppose, if possible, they share some prime divisor $p$. By Cauchy's theorem, $x^p = e$ for some $x$ in $G \setminus \{e\}$.

Consider the trivial subgroup $H = \{e\}$. Since $k$ is divisible by $p$, it follows that $x^k = e \in H$, so $x \in H = \{e\}$; that is, $x = e$, contradicting the fact that $x$ lies in $G \setminus \{e\}$. Consequently, $k$ is indeed coprime to $n$.