Subgroups of free groups which avoid conjugacy classes Let $G = (\mathbb Z/2\mathbb Z)^{\ast m}$ be a free product of some groups of order $2$.  Let $\alpha_1,\ldots,\alpha_m$ be the generators.
Can I find a free, nonabelian subgroup of $G$ that has no nontrivial elements conjugate to any $\alpha_i \alpha_j$? How can I prove it?
 A: I'll answer the stronger version of your question where the set of words $\alpha_i \alpha_j$ is replaced by any finite subset $A \subset G$.
This is not possible if $m=1$ because $G$ is finite in that case and so has no free, nonabelian subgroup.
It is also not possible if $m=2$ because $G$ is the infinite dihedral group which has an index 2 abelian subgroup (in fact cyclic) and therefore has no free nonabelian subgroup.
So we need to assume $m \ge 3$.
Every element of $G$ is expressed uniquely as a "reduced word" meaning a sequence of the form $\alpha_{i_1} .... \alpha_{i_k}$ in which any two consecutive letters $\alpha_{i_j} \alpha_{i_{j+1}}$ are unequal. The identity corresponds to the empty word with $k=0$.
Every conjugacy class in $G$ has a representative which is expressed semi-uniquely as a "cyclically reduced word" meaning that it is reduced and $b_{i_m}, b_{i_1}$ are unequal; by "semi-unique" I mean that such a representative of the conjugacy class is unique up to cyclic permutation of the word.
Okay, so, the first step is to express the conjugacy class of each element of $A$ as a cyclically reduced word, and then take $k$ to be the maximum length of those words.
Here's a particularly simple construction if $m \ge 4$.
Pick distinct reduced words $w,v$ of length $>k$ such that the starting and ending letters of $w$ and $v$ are 4 different letters, for example:
$$w = (\alpha_1 \alpha_2)^k \alpha_3
$$
$$v = (\alpha_2 \alpha_3)^k \alpha_4
$$
It follows that every nontrivial reduced word in the letters $w$ and $v$, after substitution, becomes a cyclically reduced word in the letters $\alpha_1,\ldots,\alpha_4$, and furthermore it has length $\ge k$. For example
$$w^{-1} v = \alpha_3 (\alpha_2 \alpha_1)^k (\alpha_2 \alpha_3)^k \alpha_4
$$
Therefore, the group $\langle w,v \rangle$ is a rank 2 free group and every nontrivial element in it is cyclically reduced of length $> k$, hence is not conjugate to any element of the set $A$.
If $m=3$ it's not possible to choose $w,v$ in such a simplistic manner. But one can choose the $w,v$ to be long reduced words (of length $\ge k + 4$) in the letters $\alpha_1,\ldots,\alpha_3$ so that each of the concatenations $ww$, $vv$, $wv$, $wv^{-1}$, $vw$, $vw^{-1}$ produces a word in $\alpha_1,\alpha_2,\alpha_3$ with short cancellation (at most $2$ letters are cancelled). It then follows that each reduced word in the symbols $w,v$ evaluates to a word in the letters $\alpha_1,\alpha_2,\alpha_3$ whose cyclic reduction has length $\ge k+2$, and so is nontrivial and not conjugate to any element of $A$.
A: As Lee Mosher says this is not possible if $m \le 2$. If $m \ge 3$ we can argue a bit differently as follows. $G$ is residually finite (proof), so we can find a normal subgroup $N$ of finite index not containing any finite set of non-identity elements, in particular the set $\{ \alpha_i \alpha_j \}$. Since $N$ is normal it doesn't contain any conjugates of those elements either. It remains to show that $N$ contains a free nonabelian subgroup.
By the Kurosh subgroup theorem, $N$ is a free product of finitely many copies of $\mathbb{Z}$ and $\mathbb{Z}/2$. It has a natural map to the direct product of the copies of $\mathbb{Z}/2$ only, whose kernel is a normal subgroup $N'$ of finite index which is free (this follows from some stuff about coverings of graphs of groups, or equivalently a slightly more precise form of the Kurosh subgroup theorem). Since $N'$ has finite index in $G$ it must be nonabelian (this is where we use the hypothesis that $m \ge 3$), e.g. because $G$ is not virtually abelian, or using the fact that the orbifold Euler characteristic $\chi(G) = \frac{m}{2} - (m-1) = 1 - \frac{m}{2}$ is negative.
