## Injectivity of a group homomorphism with domain a free product

Hello! Let $K = G\ast_U H$ be the free product of the groups $G$ and $H$ amalgamated along the common subgroup $U$. I have a homomorphism of groups $f:K\to K'$ defined using the universal property of free products (i.e., it is the unique map determined by two homomorphisms $j_1 : G\to K'$ and $j_2 : H\to K'$ that agree on $U$). I know that both $j_1$ and $j_2$ are injective and that the intersection of their images in $K'$ is just the image of $U$.

(1) Is it sufficient to conclude that the universal morphism $f$ is injective on $K$?

(2) If, as I suspect, the answer is no, is there some criterion to derive the injectivity of $f$? For example, let $A$ and $B$ denote the images of $j_1$ and $j_2$. Is it sufficient to prove that for each $h\in H\setminus U$ the double coset $A j_2(h) A$ is disjoint from $B$, plus the similar statement exchanging the role of the two subgroups?

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If $G=H=\mathbb Z$, $U=(e)$, $K′=\mathbb Z\oplus\mathbb Z$ and $j_1$ and $j_2$ are the inclusions on the two summands, you get a counterexample to (1) – Mariano Suárez-Alvarez Aug 13 at 19:35
It basically boils down to the normal forms theorem. I don't think you can do better. – Benjamin Steinberg Aug 13 at 19:43

If you have a homomorphism to $K'$ where $K'$ acts on a set $X$, then you could use ping-pong arguments to verify injectivity. In your setting, this is called 1-st Maskit combination theorem, you can find some examples in Maskit's book "Kleinian groups". If $K=G_1\star G_2$, then ping-poing setting is a pair of subsets $A_1, A_2\subset X$ so that $A_1\cap A_2\ne \emptyset$, and $\forall g\in G_i\setminus 1$, $gA_i\subset A_i^c\cap A_{i+1}$, where $A_i^c=X\setminus A_i$, $i=1,2$ mod 2. Under this condition, you do get a faithful action of $K$ on $X$.

Similar thing works for more general amalgams $K=G_1 \star_H G_2$, but the ping-pong condition becomes more technical.

Edit: Here is the ping-pong setting for general amalgams (Maskit's 1st combination theorem). You have an action of a group $G=G_1\star_H G_2$ on a set $X$, where each $G_i$ acts faithfully and you are trying to determine if the action of $G$ is faithful. A subset $X'\subset X$ (which need not be $G$-invariant) is partitioned in three (disjoint, nonempty) subsets $U_1, U_2, S$. You should think of two complementary components ("open half-spaces" or "halfs of the ping-pong table") to a hypersurface $S$ ("the net on the ping-pong table") in a manifold $X$. If $X$ is a complex manifold, you are in trouble since such $X$ does not come equipped with natural real hypersurfaces. One frequently uses a bisector for a suitably chosen metric on $X$ to define such $S$. Define $V_i=U_i\cup S$, $i=1,2$ ("closed half-spaces").

The ping-pong assumption then is that each $V_i$ ($i=1,2$) is strictly invariant under the subgroup $H\le G_i$, which means that $h(V_i)=V_i$ for every $h\in H$, and $g(V_i)\subset U_{3-i}, \forall g\in G_i\setminus H$. (The terminology strictly invariant is Maskit's. Notice that $U$'s and $V$'s both appear in the definition.) Examples could be found in Maskit's book.

If you think of $g\in G_i\setminus H$ as a player on the $i$-th team, then it sends the ball (point in $V_i$) from its side of the table (including the net) to the opponent's side $U_{3-i}$.

If such partition of $X$ exists, then the action of $G$ on $X$ is faithful. The proof is a direct application of the normal form for the elements of $G$. Namely, write $g\in G$ as the product, say, $g=h_0 g_1 h_1 g_2 h_2 ... g_{n}$ where each $h_j\in H$, $g_{2m+1}\in G_1\setminus H, g_{2m}\in G_2\setminus H$. Now, pick a point $x\in S$ and inductively apply each letter in this product to $x$ and check where the image lands: You will see that it never returns to $S$, unless $g=h_0\in H$.

There is a similar combination theorem for HNN extensions ("2nd combination theorem").

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 Thanks! I will check this out. Indeed I have an action of $K'$ on a set (even a variety), although it is not exactly easy to deal with. – Alberto Aug 14 at 13:15

To expand on Mariano's example and obtain counterexamples to (2), consider the case that $F_2=K=G*H$ is a rank 2 free group. Write a long rather random element of $F_2$ using the generators $G=\langle g \rangle$ and $H=\langle h \rangle$. Take $N$ to be the normal closure of that element. Unless you are very unlucky, $N$ will contain no elements of any of the forms matching elements of $G$, $H$, and the double cosets you mention. Although I haven't checked in detail, I would guess that taking $N$ to be the normal closure of $g^{35783} h^5g^2h^{-3}g^{-732}h^{42}ghghghgh$ will work, despite not being all that random.

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