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First of all, I have read this similar question and am satisfied that the answers there prove the result I am interested in. That being said, I'm more interested in this particular approach than in the actual result, so I think it's reasonable to post this as a new question, even if the approach covered there can be considered a "better" answer to the underlying question.

Fix the notation $$F_k := \textrm{ the free group on } k \textrm{ generators}.$$ Let $n,m \in \mathbb{Z}^+$ and furthermore let $n > 1$. I claim that there exists a subgroup $G$ of $F_n$ with $$G \cong F_m.$$ To "prove" this statement (actually, I believe the argument is only a heuristic):

First, the statement is trivial if $m < n$, so assume $m > n$. For the cases $n = 2, 3$, to demonstrate that $F_n$ contains an isomorphic copy of a "larger" free group, write $$F_2 = \langle a,b \rangle; \ \ \ G = \langle aa,bb,ab \rangle \subset F_2; \ \ \ G \cong F_3 $$ and $$F_3 = \langle a,b,c \rangle; \ \ \ G = \langle ab,ac,bc,aa \rangle \subset F_3; \ \ \ G \cong F_4. $$

Finally, to demonstrate that when $n \geq 4$, $F_n$ contains an isomorphic copy of a "larger" free group, write $$F_n = \langle g_1, \dots, g_n \rangle.$$

We use the observation that there are no relations between the elements of the set $$B := \{g_i g_j : 1 \leq i < j \leq n \}.$$ This set has cardinality ${n \choose 2} > n$, so the group $$G = \langle B \rangle \subset F_n$$ is isomorphic to $F_{n\choose 2}$.

Now, I'm not sure if the observation in the last few lines is actually true, and I would really love a counterexample if one exists or a hint of how I might begin a proof of it. But, supposing it holds true, then this line of reasoning apparently tells us how to write a proof of the original statement for any particular $n >1$, $m >n$. What I'm not sure of is how to write a rigorous argument that encapsulates every case without making use of the statement "and now if we wanted to we could..." or some such.

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do you know how to prove that $F_3$ embeds in $F_2$? – janmarqz Jan 9 '14 at 4:51
There is an algorithmic process known as Nielsen reduction that can be used to decide whether any given subset of a free group freely generates a subgroup. – Derek Holt Jan 9 '14 at 8:59
Stallings foldings also work, which is a more visual method than Nielsen reduction. – user1729 Jan 9 '14 at 17:00
up vote 4 down vote accepted

In the case $n=4$ let the generators of $F_n = \langle g_1, \dots, g_n \rangle$ be for simplicity denoted $F_4 = \langle a,b,c,d \rangle$ so that your set $B$ contains all six two-letter products with the letters in ascending alphabetic order. Then from $ad,bd,bc$ we can express $ac$ as $$(ad)(bd)^{-1}(bc)=add^{-1}b^{-1}bc=ac.$$ So it seems the products in $B$ may have relations between them.

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Thanks--I don't know how I managed to miss this. It was rather late last night. After a little thought I believe I think I can extend your counterexample for this case to at least one counterexample for each case $n\geq 4$, so I'm satisfied that a different approach to proving this result is needed. – Nick Jan 9 '14 at 12:11
Any sequence of four generators $a,b,c,d$ among the initial list of generators which are in increasing lexicographical order, which is used to set up the set $B,$ will give this counterexample. IMO it wouold be interesting if one could get all the relations satisfied by the sets $B$ this way. That would give a description of a class of finitely generated groups, each probably infinite. – coffeemath Jan 9 '14 at 18:27

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