Kurosh subgroup theorem

Remarks

One proof uses Bass–Serre theory. The free product $G$ acts on a tree with trivial edge stabilisers, and the vertex stabilisers are the conjugates of the factors $A_i$. The subgroup $H$ acts on the same tree, so it is the fundamental group of a graph of groups with trivial edge groups. Such a group is the free product of its vertex groups and of a free group, the fundamental group of the underlying graph. A topological proof uses covering spaces of a wedge of spaces whose fundamental groups are the $A_i$.

Example: if every $A_i$ is infinite cyclic, then $G$ is free. Each subgroup of $\Z$ is trivial or infinite cyclic, so each $g_j B_j g_j^{-1}$ is free, and $H$ is free. This recovers the Nielsen–Schreier theorem. Another example: in $\Z/2 * \Z/3$, which is isomorphic to $PSL(2, \Z)$, every subgroup without elements of finite order is free.

Going deeper

For a student

The theorem is the main structure result for subgroups of free products. It generalises the Nielsen–Schreier theorem, which says that subgroups of free groups are free. It implies that a subgroup that meets every conjugate of every factor only in the identity is free. In Bass–Serre theory it is the special case of trees with trivial edge stabilisers; the general case describes subgroups of amalgamated products and HNN extensions.