A user asked the following question. It was closed as off-topic, or rather as missing context, but it seems the context close reason doesn't exist, so off-topic was chosen. Here it is:
I am having trouble with this group theory question. Could some one solve it for me?
Let $G$ be an abelian group, and let $a\in G$. For $n\geq1$, let
$$G[n; a] = \{x\in G : x^n = a\}.$$
a. Show that $G[n; a]$ is either empty or equal to $\alpha G[n] = \{αg : g\in G[n]\}$, for some $\alpha\in G$ where $G[n] = \{x\in G : x^n = 1\}$
b. If $G$ is cyclic of order $m$, prove that:
$$|G[n; a]| = \begin{cases} \operatorname{gcd}(n, m) & \text{if} \operatorname{ord}(a)\mid\frac{m}{\operatorname{gcd}(n,m)}\\ 0 & \text{otherwise} \end{cases}.$$
Claim a. is rather easy since if $\alpha^n=a$, then $(\alpha x)^n=\alpha^nx^n=\alpha^n=a$ whenever $x^n=a$, meaning $\alpha G[x]\subseteq G[n;a]$, and on the other hand if $\beta^n=a$ then $(\alpha^{-1}\beta)^n=\alpha^{-n}\beta^n=a^{-1}a=1$ so $\alpha^{-1}\beta\in G[x]$ proving the other inclusion.
Claim b. is a little harder. I tried something, as the comments to the other question, but it is rather boring and hands-on. I was wondering if there is a not-too-calculative way to prove this. Is there? And if not, can you post one anyway, just to have one?
