Let $n, k\in\Bbb{N},n\ge 2, k\ge 2$ and $(G,\cdot)$ be of order $n$. Prove $(n,k)=1 \iff$ for all $H\le G$ we have $H=\{x\in G \mid x^k\in H\}$ 
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 $H=\{x\in G \mid x^k\in H\}$.       

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.
 A: The problem can be restated as follows:

Let $G$ be  group of order $n$. For $H$ a subgroup of $G$ and $k \in \mathbb N$, let $H_k =\{x\in G : x^k\in H\}$. Then $\gcd(n,k)=1$ iff $H=H_k$ for all subgroups $H$ of $G$.

Here is a proof.
$\Rightarrow$: 
If $\gcd(n,k)=1$, then write $nu+kv=1$. Let $x \in G$ such that $x^k \in H$. Then $x=(x^k)^v \in H$. This proves that $H_k \subseteq H$. Since clearly $H \subseteq H_k$, we have $H=H_k$.
$\Leftarrow$: 
If $\gcd(n,k)>1$, then let $p$ be a prime divisor of $\gcd(n,k)$. By Cauchy's theorem, there is $g \in G$ of order $p$. Let $E$ be the trivial subgroup of $G$. Then $g \in E_p \subseteq E_k$ and so $E_k \ne E$.
A: I'm not sure why lhf's answer has not been accepted--it is correct and concise. 
Perhaps the issue is that this answer does not address whether the proposed idea for proving the "$\Rightarrow$" portion is enough to yield the desired conclusion? 
If so, I will oblige$\ldots$ I do not believe that the given observation is quite enough, though it is a good idea. (I follow lhf in defining $H_k := \{x \in G \,:\, x^k \in H\}$ and restating the second half of the equivalence as "for all subgroups $H$ of $G$ we have $H = H_k$"). It seems to me that you are constructing a bijection between $H$ and $H \cap H_k$. The existence of such a bijection demonstrates that $H \subseteq H_k$, but I cannot see how it would imply $H_k \subseteq H$. However, your logic can be extended to imply the existence of a bijection $f: G \rightarrow G$ defined by $f(x) = x^k$. In this case $f^{-1}(H) = H_k$ (whereas before $f^{-1}(H) = H_k \cap H$), allowing you to conclude both $H \subseteq H_k$ and $|H| = |H_k|$.
