Romanian Mathematical Olympiad

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 $n$.