For a subgroup $K \leq H$, we say $K$ is a free factor of $H$ if $H$ can be written as the free product $K * C$ for some $C \leq H$, i.e. if we have the presentations $K = \langle S_K \mid R_K \rangle$, $C = \langle S_C \mid R_C \rangle$, then $H = \langle S_K, S_C \mid R_K, R_C \rangle$.
We say that $K$ is malnormal in $H$ if $\forall g \in H \setminus K$, we have that $g^{-1} K g \cap K = 1$.
In a paper I am reading, it is stated that every free factor is trivially malnormal. I've been trying to convince myself that this is true for quite some time and have not been able to. I would appreciate it if someone could explain to me why this is true.
Here's a geometric proof assuming basis Bass-Serre theory.
Write $H=K\ast L$. By Bass-Serre theory, there exists a action of $H$ on a tree with trivial edge stabilizers, with a vertex $v$ whose stabilizer is $H_v=K$.
For $h\in H$, the stabilizer of $hv$ is $hH_vh^{-1}=hKh^{-1}$. If $h\notin K$, $hv\neq v$ and $H\cap hKh^{-1}$ fixes the whole segment $[v,hv]$. Since edge stabilizers are trivial, it follows that $H\cap hKh^{-1}=\{1\}$.