set of nth-power elements of a finite abelian group Let $G$ be a finite abelian group, and let $H$ be a subset of $G$ defined by for each $n\in\mathbb{N}$ $$H=\{x^{n}\,:\,x\in G\}.$$
I know that the $H$ is indeed a subgroup of $G$.
So, by Lagrange's theorem, the order of $G$ is divisible by $H$.
Is it possible to compute the order of $H$ if the order of $G$ is given?
(The case what i'm dealing with $|G|=525$ and $n=21$. I wonder if there is a general way to compute the order of $H$.)
Give some advice or any hint. Thank you!
 A: Using fundamental theorem of abelian groups $G$ should be isomorphic to one of these groups $\mathbb{Z}_3\times \mathbb{Z}_7\times \mathbb{Z}_5 \times \mathbb{Z}_5$, $\mathbb{Z}_3\times \mathbb{Z}_7\times \mathbb{Z}_{25}$. It can be seen that $21=3 \times 7$ does not affect any any element of order $5$ or of order $25$. Therefore if $x\in G$ such that $\mid x \mid =5$ or $25$ then $\mid x^{21} \mid= \mid x\mid$. So $\mid H\mid$ is multiple of $25$. Now if $x\in G$ such that $\mid x \mid=3$ or  $\mid x \mid=7$ then $ x^{21}=(x^{3})^7=(x^7)^3=e$. Therefore the subgroup $H$ only contains elements of order $5$ or of order $25$. Hence order of $H$ is $25$.  
A: The answer is no in general. For example, if $|G|=4$ and $n=2$, then $H$ could have order $1$ or $2$.
In the particular case you are interested in though, it's not too hard to see that $G=G_5\times C_{21}$, where $G_5$ is the Sylow $5$-subgroup of $G$. It's then not hard to see that $H=G_5$ (more or less as in MANI's answer, if it was written carefully), so $|H|=25$.
More generally, this argument works whenever $G$ is an abelian group (or even nilpotent), $n$ divides $|G|$ but is coprime to $|G|/n$.
