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Let $G_1,G_2$ be groups and $H_1$ and $H_2$ be free subgroups of $G_1$ and $G_2$ respectively. Let $$ \phi:G_1*G_2 \to G_1 \times G_2 $$ be the natural surjective morphism (sending a word to the pair consisting in the product of its letters in $G_1$ and $G_2$ respectively).
Question: Is the preimage $\phi^{-1}(H_1 \times H_2)$ free?
Note that in the case where $H_1,H_2$ are trivial, this is true (Proven for example in Serre's book “Trees”, page 6).
The answer is yes, as pointed out by Derek Holt in a comment:
Let $H:= \phi^{-1}(H_1 \times H_2)$, and let $\phi_1, \phi_2$ denote the components of $\phi$. By Kurosh Subgroup Theorem, $H$ may be written: $$ H = F(X) * (*_i u_iK_{1,i}u_i^{-1}) * (*_j w_jK_{2,j}w_j^{-1}) $$ for a number of subgroups $K_{1,i}$ of $G_1$ and $K_{2,j}$ of $G_2$, a subset $X$ of $G_1 * G_2$ and words $u_i,w_j$ of $G_1*G_2$. If $H$ were not free, then some of the $K_{1,i}, K_{2,j}$ would not be free, say $K:=K_{1,i}$; assume this is the case. By construction, $u_iKu_i^{-1}$ is then a non-free subgroup of $H$, and $$ \phi_1(u_iKu_i^{-1}) = \phi_1(u_i)\phi_1(K)\phi_1(u_i)^{-1} $$ is a subgroup of $H_1$, conjugate to $\phi_1(K)$. On $G_1$, viewed as a subgroup of $G_1*G_2$, $\phi_1$ is injective, so that $\phi_1(K) \cong K$ is not free, and $\phi_1(u_i)\phi_1(K)\phi_1(u_i)^{-1}$ isn't either. Since $H_1$ contains a non-free subgroup, it can't be free, and we have reached a contradiction.
