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I am not so familiar with free groups so I want to study those groups. For example the free product is in general not a free group: If we consider the free product $\mathbb{Z}_2 \ast \mathbb{Z}_2 $, then this product is isomorphic to $ D_{\infty}$.

My Question is: Is the free group $\mathbb{F}_2$ a subgroup of $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$? If yes, how can you see this?

Sorry for my bad english.

Thanks in advance!

Seirios
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TBH
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    If the generators of the three free factors are $a,b,c$ then the subgroup $\langle ab,bc\rangle$ is free of index $2$. – Derek Holt Dec 05 '16 at 21:22
  • You can use the (details behind the proof of the) Kurosh subgroup theorem. – Qiaochu Yuan Dec 05 '16 at 21:40
  • Even more is true: If $G=G_1 * G_2$ is a free product of nontrivial groups, then, unless $G_1\cong G_2\cong Z_2$, the group $G$ contains $F_2$. – Moishe Kohan Dec 06 '16 at 01:02
  • A related question: http://math.stackexchange.com/questions/279385/when-is-g-ast-h-solvable – Seirios Dec 06 '16 at 16:10
  • I should add to my comment that, if you are familiar with the Reidemeister-Schreier method for calculating subgroup presentations, then the proof that $\langle ab,bc \rangle$ is free is fairly trivial. – Derek Holt Dec 06 '16 at 16:14
  • @DerekHolt Can you write more about the Reidemeister-Schreier method ? And is there a straightforward way to show that $<ab,bc>$ is isomorphic to $\mathbb{F}_2$ ? I have problems with the induction on the length of a reduced word (we only have to show injectivity) – TBH Dec 08 '16 at 21:22

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