I already asked this question. However, it was closed off. I now have inserted by attempt at the solution:

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!

(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

  • $\begingroup$ The original question has been answered: math.stackexchange.com/a/1543688/589. $\endgroup$ – lhf Dec 3 '15 at 1:06
  • $\begingroup$ Yes, I appreciate that. I just wanted to make sure I was correct with my understanding of your answer and attempted to answer it fully to make sure I was on the right track @lhf $\endgroup$ – queence Dec 3 '15 at 1:18
  • $\begingroup$ On this site it is better to use standard math notation, not computer code.Many of us do not know what >:= means. $\endgroup$ – DanielWainfleet Dec 3 '15 at 1:31
  • $\begingroup$ sorry, it was not supposed to be ">:=". It was due to the effect of putting it in the yellow box. I have fixed it. thank you! @user254665 $\endgroup$ – queence Dec 3 '15 at 1:48

(I)...In general $F(x)=x^n$ is neither a surjection nor an injection. Consider a two-element group and $n=2$...(II). Observe that $G[n]\ne \phi$ because $1\in G[n]$. Now suppose $G[n,a]\ne \phi$. Let $x\in G[n,a].$Since $G$ is commutative, we have $x G[n]\subset G[n,a].$ For the reverse inclusion, if $y\in G[n,a]$, then $ x^{-1}y\in G[n]$ (because $G$ is Abelian) so $y=x(x^{-1}y)\in x G[n].$


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