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I am thinking that if a free group $F_n$ on $n$ generators contains an infinite cyclic normal subgroup then $n$ must be equal to 1 (i.e. $F_n$ was actually isomorphic to $\mathbb{Z}$ itself). Is this correct?

My reasoning would be that: if I have $n\geq 2$, for example take $F_2 = \langle a,b | \ \rangle$, then any infinite cyclic subgroup should be a subgroup of either $\langle a \rangle \cong \mathbb{Z} $ or $\langle b \rangle \cong \mathbb{Z} $, but this would no longer be normal.

stoneaa
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  • Strongly related: https://math.stackexchange.com/questions/168913/normal-subgroups-of-free-groups-finitely-generated-implies-finite-index – Eric Towers Mar 04 '25 at 04:30
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    Why isn't $\langle ab \rangle$ an infinite cyclic subgroup of $F_n$ for $n \geq 2$? In fact, most words in $F_n$ are generators of an infinite cyclic subgroup. The hard part is finding one that's normal... – Eric Towers Mar 04 '25 at 04:32

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