I am reading through this proof that if $H$ is a non-trivial group, then the normalizer of $H$ in the free product $G:=H \ast K$ is equals $H$ (i.e., is trivial). Most of the proof seems to generalize to amalgamated free products. There's just one possibly troublesome step at the very end. Basically, the proof goes as follows.
Assume by way of contradiction that $H$ is not self-normalizing, so there exists $g \in N_{G}(H) \setminus H$. Without loss of generality, assume the first and last letter of $g$ come from the $K$ factor. Now, choose any non-trivial element $h \in H$. Then $ghg^{-1}$ is also a reduced word of length greater than one, so $ghg^{-1} \notin H$, which contradicts the fact that $g$ normalizes $H$.
My question is:
Does this fact generalize to amalgamated free products, or is there a nice class of examples where we have this? That is, if $H$ and $K$ are amalgamated over a group $A$ (not sure if that's the correct vernacular), is $H$ self-normalizing in $H \ast_{A} K$?
For my purposes, I don't really care $A$ and $K$ are (they can be fixed), but I would like $H$ to be more or less arbitrary (or as arbitrary as possible if that makes sense).
The fact that we can just choose any non-trivial $h \in H$ and we get the desired contradiction is interesting. If it's not generally true that $H$ is self-normalizing in $H \ast_{A} K$, is it possible there is a "nice" and not too restrictive class of $H$'s where we can show that $H$ is self-normalizing in $H \ast_{A} K$.
As mentioned by Derek Holt, you just have to be a bit more careful because the "normal form" for $H*_AK$ is different.
A common way of defining the normal form is the following: pick a set of left coset representatives $\{h_i\}_{i\in I}$ for $A$ in $H$, with $e_H$ the representative of $H$; and a set of left coset representatives $\{k_j\}_{j\in J}$ for $A$ in $K$, with the representative of $A$ being $1_K$.
Then the elements of $H*_AK$ have can be expressed uniquely in the form $$h_{i_1}k_{i_1}\cdots h_{i_r}k_{i_r}a$$ where $a\in A$, and none of the $h_{i_s}$ or $k_{j_t}$ are equal to $1$, except perhaps for $h_{i_1}$ and/or $k_{i_r}$.
To explain how to multiply two such elements let's consider the case of an element of the form $h_{i_1}k_{j_1}a$ times an element of the form $k_{j_2}h_{i_3}k_{j_3}a'$. Write out the element: $$h_{i_1}k_{j_1}ak_{j_2}h_{i_3}k_{j_3}a'.$$ Now we shift $a$ to the "end". The element $k_{j_1}ak_{j_2}$ lies in $K$, so it can be written uniquely as $k_{u_1}a_1$ for some $u_1\in J$, $a_1\in A$. This gives $$h_{i_1}k_{u_1}a_1h_{i_3}k_{j_3}a'.$$ If $k_{u_1}$ is not trivial, then rewrite $a_1h_{i_3}$ as $h_{v_1}a_2$ and replace it in the expression. If $k_{u_1}=1$, then rewrite $h_{i_1}a_1h_{i_3}$ in the form $h_{v_1}a_2$ and replace it in the expression. Etc. Continue this way until the $a$ has been "shifted" all the way to the right, and left of it you have an alternating list of coset representatives in $H$ and in $K$.
Then in the final part of your argument you can take any $h\in H$, and an element $x$ not in $H$. Assuming it normalizes $H$, you can assume that it has the form $k_{j_1}\cdots h_{i_r}k_{j_r}$ (by left and right multiplying by an element of $H$, which certainly normalizes), and $k_{j_1},k_{j_r}\neq 1$, and write $h=h_ua$. Then $xhx^{-1}$ will have normal form that starts with $k_{j_1}$ and so does not lie in $H$.