Is the map $G*H \to G \times H$ injective?

Question

I am reading Hatcher's book on Algebraic topology. Here is a paragraph about free product of group.

I think the last line should not be "a surjective homomorphism", but an injective one, since the free product group is smaller than direct porduct group. Am I correct?

Also, is the homomorphism $G*H \to G \times H$ simply inclusion homomorphism? I treat $G*H$ as a subgroup of $G \times H$

Answer 1

No it is not, you have a surjection from $\mathbb{Z}*\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z} $ whose kernel is the normal subgroup generated by $xyx^{-1}y^{-1}, x,y \in \mathbb{Z}.$

Answer 2

No. As you say, the free product of two groups is not "smaller" than the direct product, an easy counter example is $\Bbb{Z}/2\Bbb{Z} \ast \Bbb{Z}/2\Bbb{Z}$ is an infinite group, where $\Bbb{Z}/2\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z}$ has only 4 elements.