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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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    my goodness, well this is a sur$p$rise. didn't expect that subgroups can have larger rank that the whole group. Thank you.2011-04-24

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Consider the lovely example that arises from studying the usual presentation of $S_3$, namely $$. 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.