Index of subgroups of "$k$-powers". Let $G$ be a cyclic group of order $n$. Let $G_k$ the subgroup
$$G_k=\left\{x^k: x\in G\right\}.$$
Is it true that $[G:G_k]\in\{1,k\}$?
If $n=p-1$ and $k=2$ this is true and I used many times in some number theory exercises. How much can I generalize this thing?
What if $G$ is any (maybe abelian) group?
 A: No: For $G=C_{12}$ we have $[G:G_8]=3$.
In general, for $G=C_{n}$ we have $[G:G_k]=\dfrac{n}{\gcd(n,k)}$.
A: Here is a similar one using lhf hint:
Take $G=\Bbb Z_{6}$ and $$G_4=\{x^4=4x=x+x+x+x:x \in \Bbb Z_{6}\}=\{0,4,2\}$$ and $[G:G_4]=2\neq1,4$

Your result is clearly true if $k=2$, since $$\text{the number of squares in a cyclic group of order $n$}=\begin{cases} n&\text{if}\;n \;\text{is odd} \\\\\frac{n}{2} &\;n \;\text{is even}\end{cases}$$
A: If $G$ is cyclic of order $n$, say $G=\langle a \rangle$ then the subgroup of $m$-powers is also cyclic generated by $a^m$ and it can be shown that the order of $a^m$ is $\frac{n}{(m,n)}$: if $a^{km}=1_G$ then $n|km$; after simplifying this divisibility relation by $(m,n)$, since $\left(\frac{n}{(m,n)}, \frac{m}{(m,n)}\right)=1$ we infer right away that $\frac{n}{(m,n)}|k$. 
Hence, $|G:\langle a^m \rangle|=(m,n)$.
In general, for arbitrary $n \in \mathbb{N}^{*}$ and arbitrary abelian group $(G, +)$ the index $|G:nG|$ can be arbitrarily large: consider for instance a direct sum $G=(\mathbb{Z}/2n\mathbb{Z})^{(T)}$, for arbitrary set $T$. You will have
$$G/nG \approx (\mathbb{Z}/2\mathbb{Z})^{(T)}$$
and thus $|G:nG|=|T|$, when $T$ is infinite.
