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As the title suggests, my question is does $\mathbb{Z}* \mathbb{Z} /N = \mathbb{Z} * \mathbb{Z}$ imply that $N$ is trivial? Here, $\mathbb{Z} * \mathbb{Z}$ is the free product of $\mathbb{Z}$ and $\mathbb{Z}$ and $N$ is some normal subgroup of $\mathbb{Z}* \mathbb{Z}$. My intuition tells me that the answer is yes, despite this not holding for other groups; a quick example is $\mathbb{S}^1 = \{ e^{i\theta} \in \mathbb{C} \}$, since $\mathbb{S}^1/\{\pm 1\} = \mathbb{S}^1 $.

An equivalent problem would be, given a surjective epimorphism $\varphi: \mathbb{Z} * \mathbb{Z} \to \mathbb{Z}*\mathbb{Z}$, does $\varphi$ have to be injective as well?

Marcus M
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2 Answers2

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Yes, $\mathbb{Z}*\mathbb{Z}$ is just the $2$-generator free group which is Hopfian http://en.wikipedia.org/wiki/Hopfian_group.

user 59363
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Finitely generated free groups are always Hopfian, ie. every epimorphism $G \to G$ is an isomorphism. This is precisely what you want to prove for $\mathbb{Z} * \mathbb{Z}$, the free group on two generators.

The proof can be found here (and on the linked pages). The idea is that:

  • a free group is always residually finite (ie. $\forall g \in G \setminus \{e\}$, there is a normal subgroup $N$ of $G$ of finite index not containing $g$); this can be proven relatively easily using the description of $g$ as a word on the generators and building a ad-hoc nontrivial morphism $f : G \to K$, where $K$ is finite and $f(g) \neq e$.

  • a residually finite, finitely generated group is Hopfian: given an epimorphism $f : G \to G$, if $f(g) = e$ ($g \neq e$) you can use the residually finite property to find a nontrivial morphism $h : G \to K$ with $K$ finite and $h(g) \neq e$; then the $h \circ f^n$ are all pairwise distinct morphisms, but there's only a finite number of morphisms $G \to K$ (because $G$ is finitely generated).

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Najib Idrissi
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