We were given questions to study for our exam coming up. We have not covered much of this topic, so any help would be greatly appreciated!
Let $G$ be an abelian group, and let $a\in G$. For $n≥1$, let $G[n:a] := \{x\in G:x^n =a\}$.
(a) Show that $G[n: a]$ is either empty or equal to $αG[n] := \{αg : g \in G[n]\}$, for some $α∈G$. (Recall: $G[n]:=\{x\in G:x^n =1\}$.)
(b) If $G$ is cyclic of order $m$, prove that:
$|G[n:a]| = (n,m)$ if $\text{Ord}(a)\mid(m/(n, m))$, OR $0$ otherwise.
Thank you for all your help!
EDIT:
(a) We want to show that G[n: a] is either empty or equal to $αG[n] := \{αg : g \in G[n]\}$, for some $α∈G$. We first must show that the map F: G -> G, defined by F(x) = $x^n$, is a homomorphism.
Suppose there exists some x,y in G. Then we have that $F(xy) = (xy)^n = x^ny^n = F(x)F(y)$, by the properties of abelian groups. Therefore, F is surjective. Next, we must show that F is injective.
Now, suppose x is in ker(F), where $x^n=a$. By Lagrange's Theorem, it's cyclic subgroup H, is finite and must divide the order of G. Let m = |H|. Then we have $x^m=a$, similar as before.
Now, since any common factor of n and m is also a common factor of n and the order of G, in other words: m divides odd(G), then m and n must be relatively prime to one another.
Furthermore, let there exist some integers å,b in G, such that åm + bn = 1. Then we'll have the following:
$x = x^1 = x^{åm + bn} = x^{åm}x^{bn} = (x^m)^å(x^n)^b = a$.
So we have that x = a. Therefore, ker(F) = {a}, and we have that F is injective. Hence, for all g in G, there exists some x in G such that $g = F(x) = x^n$.
Can someone please let me know if I'm on the right track? thanks
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).