Decomposability of free product of groups Let $A$ be a group that can't be expressed as the free product of two nontrivial groups (that is if $A \cong C * D$ then $C$ or $D$ must be trivial).
If $T$ is a quotient of $A$ does it have the same property?
Or (since I think the answer to the previous question is no), is there some condition to impose on $A$ such that all its proper quotients can't be expressed as a nontrivial free product?
 A: If the group $A$ has Serre's property FA then neither $A$ nor any of its proper factors are free products. Groups with Serre's property FA do not decompose as a free product (with or without amalgamation) or as an HNN-extension. Moreover, their homomorphic images also have FA, and so they satisfy the conditions you are asking about.
FA stands for Fixed Arbre, which is French for Fixed Tree (Serre is French). A finitely generated group has property FA if whenever it acts on a tree without inversion$^{\dagger}$ there is a global fixed point. Every group acts on a tree, as you can take a tree with a single parent vertex and one generation of children labelled with the elements of the group, and then the group acts by right multiplication on the children but keeps the parent fixed. Clearly, the parent is always fixed - it is a global fixed point! Property FA says that whenever you act on a tree there is such a global fixed point.
The standard reference for this is Serre's book Trees.
Examples of groups with Property FA, and so examples of groups which answer your question, are finite groups, or more generally, finitely generated torsion groups (groups all of whose elements have finite order), Triangle groups (groups with presentation $\langle a, b, c; a^i, b^i, c^i, abc\rangle$) and groups with Kazhdan's Property T. Note that if a group $G$ has a finite index subgroup $H$ with FA then $G$ has FA. However, if $G$ has FA it is not necessarily true that $H$ has FA. So, closed under supergroups but not under subgroups.
So, the question on your lips should be "Trees? What do they have to do with free products?!?" Well, quite a lot actually! This is the subject of Bass-Serre theory. The point is that a free product acts on a specific tree in a specific way (as do free products with amalgamation and HNN-extensions). If $G=H\ast K$ then there are two types of vertices, those fixed by (conjugates of) $H$ and those fixed by (conjugates of) $K$. So...waving my hands...you have two vertices and an edge joining them $v_1-v_2$ where $\operatorname{Stab}(v_1)=H$ and $\operatorname{Stab}(v_2)=K$...then construct an (infinite) graph by orbiting this by the action of $G=H\ast K$. For a concrete example, take $G=C_3\ast C_4$ and take the graph $\Gamma$ to the graph where every vertex has degree three or four, and between any two vertices of order three there is one of order four, and between any two vertices of order four there is one of order three. Then let $C_3$ act on the tree by rotating a specified vertex of valency three while $C_4$ acts by rotating a specified vertex of valency four. (This is a worked example in John Meier's very readable book "graphs, groups and trees", which contains a picture of the tree in question.)
I want to add a few questions. The first two questions are not difficult, and get you thinking about property FA. I do not know the answer to the third one, but it is probably either proved or disproved in a textbook somewhere. In fact, it is probably in Trees. (I have a proof in my head, but it seems kinda dodgy, so I am unwilling to be committal about this question!)
Question 1: Prove that $\mathbb{Z}$ does not have Property FA. Conclude that every finitely generated group which maps onto $\mathbb{Z}$ does not have Property FA.
Note that $\mathbb{Z}$ does not split as a free product with amalgamation nor as an HNN-extension. Therefore, Property FA is not equivalent to splitting (although it turns out it is equivalent to (not splitting + not mapping onto $\mathbb{Z}$)).
Question 2: Show that if $G$ has property FA then every homomorphic image of $G$ has property FA.
Note that the point of this answer is that this holds! The idea of the proof is the same idea you (should have) used for the "Conclude..." part of Question 1.
Question 3: Is it true that if $H$ is a free product and $G$ maps onto $H$ then $G$ decomposes non-trivially as a free product with amalgamation? That is, does $G\twoheadrightarrow H\ast K\Rightarrow G=A\ast_B C$ hold?
Note that in Serios example, the group is a free product with amalgamation (why?).
$^{\dagger}$ Inversion: The group $G$ acting without inversion means that for all $g\in G$, $\iota(e)\cdot g=\iota(e\cdot g)$ and $\tau(e)\cdot g=\tau(e\cdot g)$, where $\iota$ means the initial and $\tau$ the terminal vertex of an edge. Basically, it means that no edge is "flipped".
A: Let $A,B,C$ be three nontrivial groups. Suppose moreover that the center of $C$ is nontrivial. Then a counterexample is given by $$G=(A \ast B) \times C.$$
Indeed, since a nontrivial free product is centerless, we deduce that $G$ is freely indecomposable (the center of $C$ is included in the center of $G$); on the other hand, $G/C \simeq A \ast B$ is clearly freely decomposable.
