malnormal subgroup of amalgamated free product

Question

Consider the amalgamated free product $\Gamma = K\ast_{H\simeq H'} L$. Let $A$ be a malnormal subgroup of K i.e, for all $k\in K\setminus A$, $k^{-1}Ak \cap A ={1}$. Is A malnormal in $\Gamma$?

I was trying to do this by writing elements in normal form but completely stuck. For example let $g\in \Gamma\setminus A$ and $m\in A$, then i was trying to prove that $g^{-1}mg$ does not belong to A. if we write $g$ in normal form in $\Gamma$ using right cosets representative of $H,H'$ in $K,L$ respectively then $g^{-1}$ will be in normal form using left cosets of $H,H'$ in $K,L$, then $g^{-1}mg$ will no longer be in a normal form using right cosets or left cosets. I don't know how to handle this. Thanks for any help.

Answer 1

Assuming that you meant what I suggested in my comment regarding malnormality of $A$ in $K$, this isn't true. The idea for a counterexample is that $A$ could have a nontrivial intersection with $H$, and $A \cap H$ could be identified with some subgroup of $H'$ that is a proper, normal subgroup $L$.

To turn that into an actual counterexample:

• Let $K$ be the rank $3$ free group $K = \langle a \rangle * \langle b\rangle * \langle c \rangle$,
• Let $A < K$ be the rank $2$ free factor $A = \langle b \rangle * \langle c \rangle$ which is malnormal in $K$,
• Let $H < A$ be the rank $1$ free factor $H = \langle c \rangle$,
• Let $L$ be the rank $2$ free abelian group $\langle c' \rangle \oplus \langle d \rangle$,
• Let $H < L$ be the rank $1$ direct factor $\langle c' \rangle$.

And now amalgamate $K$ and $L$ using an isomorphism $\langle c \rangle \approx \langle c' \rangle$.