Grushko theorem

Remarks

Part 2 follows from part 1. If $G$ is generated by $n$ elements, take $F$ free of rank $n$ and $\varphi$ onto $G$. Then $F_1$ and $F_2$ are free factors of $F$, so their ranks add up to $n$, and they map onto $A$ and $B$, so $\rank A + \rank B \le n$. The other inequality is clear, since generators of $A$ and of $B$ together generate $A * B$. Proofs of part 1 use Stallings foldings or Bass–Serre theory.

Example: $\Z/2 * \Z/3$ has rank $2$, and $\Z^2 * \Z$ has rank $3$.

The free product cannot be replaced by the direct product. The groups $\Z/2$ and $\Z/3$ have rank $1$ each, but $\Z/2 \times \Z/3 \cong \Z/6$ is cyclic, so its rank is $1$, not $2$.

Going deeper

For a student

The first part says that every way of mapping a free group onto $A * B$ splits along a free product decomposition of the free group. The rank formula is the usual form of the theorem. It implies that every nontrivial finitely generated group is a free product of finitely many freely indecomposable groups. Here a group is freely indecomposable if it is nontrivial and not a free product of two nontrivial groups. The decomposition stops because each split lowers the ranks of the factors. It fits with the Kurosh subgroup theorem as one of the basic facts about free products.