Commutator subgroup of cyclic group Find commutator subgroup of cyclic group.
I know that every cyclic group is abelian group and commutator subgroup of any abelian group is trivial group.
I just need help to prove this. 
 A: The commutator is defined as $[a,b]=aba^{-1}b^{-1}$. Since the group is cyclic, it's Abelian. So, we have that $\forall a,b: ab=ba$. This gives that $aba^{-1}b^{-1}=e$ always. So, the commutator subgroup is $e$.
A: Every cyclic group is abelian:
Indeed, if $G$ is cyclic, then there exists $a$ such that $G=\left<a\right>$.
Let $g,h\in G$; then there exists $n,m\in\mathbb Z$ such that $g=a^n$ and $h=a^m$. Therefore $$gh=(a^n)(a^m) = a^{n+m}=(a^m)(a^n) = hg$$
The commutator subgroup of every abelian group is trivial:
Indeed, the commutator subgroup is generated by elements of the form $[g,h]:=ghg^{-1}h^{-1}$ for $g,h\in G$. If $g,h$ commutes, then $ghg^{-1}h^{-1}=gg^{-1}hh^{-1}= e\cdot e = e$. Therefore, the commutator subgroup is trivial.
A: Forgive me if perhaps I misunderstood, but is there anything more to this question than:
[$g,b$]=$g^{-1}b^{-1}gb$= $c^{-i}c^{-j}c^ic^j$=$e$ and thus the the commutator subgroup is the minimum subgroup on the set {$e$}, which is the trivial group. 
Of course, it would depend on how exactly the commutator was defined to you( maybe your professor defined it through quotients, which would yield another simple solution)
