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I just got a little confused reading the formulation on wiki. Let $F_X$ denote the free group on the set $X$ and let the symbol $\leq$ denote "is subgroup of".

From what I know, the theorem reads: $H\leq F_X$ $\Rightarrow$ $\exists Y$: $H\cong F_Y$.

Is my formulation of this theorem also correct: $H\leq F_X$ $\Rightarrow$ $\exists Y\subseteq X$: $H=F_Y$

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No, certainly not. Consider $X = \{a,b\}$ and the subgroup $H \cong \mathbb{Z}$ generated by $ab$ in $F_X$. Another example is the commutator subgroup $[F_X,F_X]$, which can be shown to be free on a countably infinite set (as soon as $|X| \geq 2$), so anything can happen in terms of the size of the generating set. –  t.b. Apr 24 '11 at 16:53
    
my goodness, well this is a surprise. didn't expect that subgroups can have larger rank that the whole group. Thank you. –  Leon Lampret Apr 24 '11 at 22:11
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up vote 2 down vote accepted

Consider the lovely example that arises from studying the usual presentation of $S_3$, namely $<a,b: a^3=b^2=(ab)^2=1>$. The presentation leads to a free group on $a$ and $b$, and the subgroup of this free group `normally generated' by the three relations $a^3$, $b^2$ and $(ab)^2$ is of rank 7. For your interpretation $X= \{a,b\}$ whilst clearly your $Y$ is not a subset of that as it has 7 elements, Doh!

One of the clearest ways of looking at Neilsen-Schreier is via covering graphs. This can be found in several places. The example of $S_3$ can be found in Ronnie Brown's book: Topology and Groupoids, page 400.

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thanks, nice reference. –  Leon Lampret Apr 24 '11 at 22:35
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