Let $G$ be a group and denote the set of conjugacy classes by $C(G)$. For a subgroup $H$ of $G$, the inclusion $H\hookrightarrow G$ descends to a map $C(H)\to C(G)$, but this is generally not injective.
Question: When is it injective?
For example, it's obviously true if $G$ is abelian. In another case that $G=F_2=\langle x,y\rangle$ and $H=F_1=\langle x\rangle$ with natural inclusion $x\mapsto x$, it's also true.
I'm especially interested in the case of the free groups, and in particular finite index normal subgroups of free groups. Any suggestion or some references will be appreciated.
Best Answer
A subgroup $H\leq G$ is called malnormal if $g^{-1}Hg\cap H$ is trivial for all $g\not\in H$. This clearly implies injectivity of your map, and malnormal subgroups are well-understood for free groups. For example, maximal cyclic subgroups are malnormal.
Moreover, every finitely generated subgroup $H$ of a free group $G$ is a free factor of (hence malnormal in) a finite index subgroup $K_H\leq_fG$; this is Marshall Hall's Theorem. So all finitely generated subgroups of $G$ are somehow "close" to being malnormal in $G$.
On the other hand, the map $C(H)\to C(G)$ will always be non-injective when $G$ is non-cyclic free and $H$ is proper of finite index. The key fact here is that such a subgroup contains with (finite index) a non-trivial normal subgroup $N$ of $G$. More generally:
Theorem. Let $G$ be a non-cyclic free group, and $H$ a proper subgroup of $G$ which contains a non-trivial normal subgroup $N$ of $G$. Then the map $C(H)\to C(G)$ is non-injective.
Proof. As $N$ is normal in a non-cyclic free group, it is itself non-cyclic and free, so there exists a pair of non-commuting elements $h_1, h_2\in N\leq H$. Set $h=h_1^2h_2^2$.
Assume the map $C(H)\to C(G)$ is injective.
Consider some $g\in G\setminus H$. By normality of $N$ in $G$ we have $g^{-1}hg=h'\in N\leq H$, where $h$ is as above. As the map $C(H)\to C(G)$ is injective, there exists some $k\in H$ such that $k^{-1}hk=h'$, and so $kg^{-1}$ commutes with $h$. As we are in a free group, $\langle kg^{-1}, h\rangle$ is cyclic. By the famous result of Lyndon and Schützenberger [1], $h$ is not a proper power of any element in $G$ and so $\langle kg^{-1}, h\rangle=\langle h\rangle$. Therefore, $kg^{-1}\in\langle h\rangle\leq H$. It follows that $g\in H$, a contradiction. QED
Note. The key point in the proof is that $H^g\cap H$ contains an element $h$ which is not a proper power in $G$. The fact that $N$ is a subgroup is therefore important as it is needed to obtain such an element $h$. In particular, this means that the proof does not immediately work for $H$ such that $H^g\cap H$ is non-trivial for some/all $g\in G$.
[1] Lyndon, R. C., & Schützenberger, M. P. (1962). The equation $a^M= b^N c^P$ in a free group. Michigan Math. J, 9(4), 289-298. (link)