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Let $G=\langle g_1, g_2,\dots\rangle$ be a free group, with no relations.

Let $H$ be a subgroup of $G$, with generators $h_1,h_2,\dots\in G$. Suppose the generators have a relation, say $h_{i_1}h_{i_2}\dots h_{i_m}=1$. Since each $h_i$ can be written as a product of $g_j$, then $h_{i_1}h_{i_2}\dots h_{i_m}=1$ implies that $g_{j_1}\dots g_{j_k}=1$ for some $j_1,\dots,j_k$.

This contradicts that $G$ has no relations for its generators.

Hence $H$ is a free group.

Intuitively, this is wrong (the proof is too simple). However I can't find the mistake.

Thanks a lot for any help.

yoyostein
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    I asked a similar question here which received some wonderful answers that might help. – Santana Afton Jul 02 '18 at 03:32
  • Does your argument only prove that a subgroup $H$ of a free group $G$ is a free group, or does it prove the stronger assertion, that any generating set for $H$ is a free generating set? Does a similar argument prove that a subsemigroup of a free semigroup is a free semigroup? – bof Jul 02 '18 at 03:40
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    Your implication is incorrect. All you get is that some word in the $g_i$ must be equal to $1$, but this word might e.g. be $g_1 g_2 g_2^{-1} g_1^{-1}$ or something similar, which is equal to $1$ even though the $g_i$ are free. – Qiaochu Yuan Jul 02 '18 at 04:02
  • @QiaochuYuan Great. I understand now. – yoyostein Jul 02 '18 at 04:18

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