Let G be an abelian group, and let a∈G. For n≥1,let G[n;a] := {x∈G:x^n =a}. Show that G[n; a] is either empty or equal to αG[n] := {αg : g ∈ G[n]}... [closed]
Solution 1:
If $G$ is an abelian group, then $\phi: G \to G$ given by $\phi(x)=x^n$ is a homomorphism.
For $a\in G$, we have $G[n:a] =\phi^{-1}(a)$ and so, if not empty, it is a coset of $\ker \phi$, which is $G[n]$ in your notation. This proves (a).
Write $d=(n,m)=nu+mv$, with $u,v\in \mathbb Z$. If $x\in G[n]$, then $x^d=1$ because $x^m=1=x^n$. Conversely, if $x\in G[d]$, then $x^n=1$ because $n$ is a multiple of $d$. It follows that $G[n]=G[d]$.
If $G$ is cyclic of order $m$ and $d$ divides $m$, then $G[d]$ has order $d$. Combined with (a), this proves (b).