Both the question and your answer contain several misconceptions/errors.
Your question seems to assert that every finite group has a maximal normal subgroup of prime index. While every subgroup of prime index is necessarily maximal (this follows from the multiplicativity of the index, if $H\lt K \lt G$ then $[G:H]=[G:K][K:H]$), it is not true that every such group is necessarily normal, and it is not true that every maximal subgroup has prime index. For example, $A_5$ has a maximal subgroup of index $5$ (the subgroup fixing $5$, say), but it is not normal; and it has a maximal subgroup of order $6$ and index $10$, generated by $(123)$ and $(12)(45)$.
The statement "if you divide the order of a finite group by the smallest prime possible, that any subgroup will be normal and of prime cyclic order" is hopelessly garbled. On the basis of your comment on the post, you tried to say that if the index of a subgroup is the smallest prime that divides the order of the group then that subgroup will be normal and the quotient cyclic of prime order. That is true, but not every finite group has such a subgroup (see above: $A_5$ has no subgroup of index $2$).
The statement in the answer that "any group not composed of cyclic simple groups will not have any cyclic simple quotient" is false if you mean "a group in which not every composition factor is a cyclic simple group". It is false for $\mathbb{Z}$, for example; and for finite groups, it is false for $S_5$ (which has a quotient isomorphic to $C_2$) and more generally, any group of the form $A\times B$ where $A$ is finite abelian and $B$ is nonabelian simple. If you mean "for which every composition factor is not cyclic simple group", then that is true for finite groups but your phrasing is extremely confusing.
To your questions:
Theorem. A finite group $G$ has a quotient that is nontrivial cyclic if and only if the commutator subgroup is a proper subgroup, $G\neq [G,G]$.
Proof. A nontrivial finite abelian group has nontrivial cyclic quotients, hence nontrivial cyclic quotients of prime order. If $G=[G,G]$, then the only abelian quotient of $G$ is trivial so $G$ cannot have nontrivial cyclic quotients. If $G\neq[G,G]$, then $G/[G,G]$ is finite abelian and hence has itself a quotient which is cyclic of prime order, and therefore so does $G$ by the isomorphism theorems. $\Box$
For infinite groups the question is harder. There are groups (even abelian groups) with no subgroups of finite index. For example, the Prüfer $p$-group $\mathbb{Z}_{p^{\infty}}$ is infinite, but all proper subgroups are finite. And the additive group of rationals $\mathbb{Q}$ does not have any maximal subgroups, and in particular cannot have a subgroup of finite index. The quotients of $\mathbb{Z}_{p^{\infty}}$ are either trivial or isomorphic to $\mathbb{Z}_{p^{\infty}}$ itself, which is quasicyclic but not cyclic. No nontrivial quotient of $\mathbb{Q}$ can be nontrivial and cyclic (even infinite cyclic) because $\mathbb{Q}$ is divisible, hence so is every quotient, and no nontrivial cyclic group is divisible. Both of these groups are abelian, so the commutator test certainly does not work in general.
It does work for finitely generated groups:
Theorem. Let $G$ be a finitely generated group. Then $G$ has a quotient that is nontrivial cyclic if and only if $[G,G]\neq G$.
Proof. It follows as above once you verify that a finitely generated abelian group has such quotients, and note that a quotient of a finitely generated group is finitely generated. $\Box$
