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> Let $n, k \in \mathbb{N}, n\ge 2, k\ge 2$ and $(G,\cdot)$ be a group with $n$ elements. Prove that $(n,k)=1$ if and only if for all subgroups $H$ of $G$ we have that $H=\{x\in G \mid x^k\in H\}$. 
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I think that the last equality is pretty odd.  

For "$\implies$", I observed that for any subgroup $H$ of $G$ the function $f:H\to H, f(x)=x^k$ is a bijection because $(n,k)=1$. From here it would follow that for any $y\in H$ there exists an unique $x\in H$ such that $y=x^k$. I am not sure if this is enough to reach our conclusion in this case.