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I saw this answer and I'm thinking if with this idea we can show that $G_1 \ast G_2 \ast F$ (where $F$ is a free group of countable rank) embedds in $G_1 \ast G_2$ if $G_1$ or $G_2$ has cardinality at least $3$.

If $G_1$ has cardinality at least $3$, we can choose $a \neq b \in G_1$ non-trivial. Consider the groups $G_1$, $aG_2a^{-1}$ and $\langle bg_2b^{-1} \rangle$ for an element $g_2 \in G_2$. I think these subgroups freely generates $G_1 \ast G_2 \ast \Bbb{Z}$. Now we can apply the same decomposition on $G_1 \ast G_2 \ast \Bbb{Z}$ to get $G_1 \ast G_2 \ast \Bbb{Z} \ast \Bbb{Z}$. Inductively, I think it shows the result.

I know that it was not written formally, but I would like to know if this idea works.

Answer. Using Derek Holt's comment, we have the following:

Let $G G_1 \ast G_2$ a non-trivial product with $|G_2| \geq 3$. Choose $a \in G_1 - \{1\}$ and $b \neq c \in G_2 - \{1\}$. Consider $H_1$ the subgroup generated by $aG_2a^{-1}$, $bG_1b^{-1}$ and $\langle g \rangle$ where $g = cabac^{-1}ac^{-1}$. First, note that $$g^2 = (cabac^{-1}ac^{-1})(cabac^{-1}ac^{-1}) = cabac^{-1}a^2bac^{-1}ac^{-1}$$ so $g^2 \neq 1$ even if $a^2 = 1$ since $bc^{-1} \neq 1$. Inductively, $g^n \neq 1$ for every natural $n$. By the choices of $a,b,c$ the elements of the groups $aG_2a^{-1}$, $bG_1b^{-1}$ and $\langle g \rangle$ have no relations since there is no cancellation in its products. So, $$H_1 \simeq aG_2a^{-1} \ast bG_1b^{-1} \ast \langle g \rangle.$$ Since $aG_2a^{-1} \simeq G_2$, $bG_1b^{-1} \simeq G_1$ and $\langle g \rangle \simeq \Bbb{Z}$, $$H_1 \simeq G_1 \ast G_2 \ast \Bbb{Z}.$$ We can apply the same argument again in $H_1$ to obtain $$H_2 \simeq G_1 \ast G_2 \ast \Bbb{Z} \ast \Bbb{Z}$$ with $H_2 \hookrightarrow H_1 \hookrightarrow G$. Inductively, we have that $G$ contains a subgroup $H$ isomorphic to $G_1 \ast G_2 \ast F_n$ where $F_n$ is the free group of rank $n$. Since $F_n$ contains a free subgroup of countably rank for $n \geq 2$ (the commutator subgroup), then $G$ contains a subgroup isomorphic to $$G_1 \ast G_2 \ast F$$ where $F$ is a free goup of countably rank.

Is this correct?

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    $\begingroup$ $bg_2b^{-1}$ lies in the subgroup generated by $G_1$ and $aG_2a^{-1}$. So the subgroup they generate is contained in $G_1*G_2$. $\endgroup$
    – Arturo Magidin
    Commented Jun 1 at 3:57
  • $\begingroup$ @ArturoMagidin is there some way to get an element not generated by these subgroups? Or other way to get $G_1 \ast G_2 \ast \Bbb{Z}$? $\endgroup$
    – Greg
    Commented Jun 1 at 4:01
  • $\begingroup$ Also, we need an element of infinite order. Something like $g_1g_2$ to generates $\Bbb{Z}$. $\endgroup$
    – Greg
    Commented Jun 1 at 4:56
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    $\begingroup$ @SteveD I know the statement of the theorem, but not the proof. Is this proof helpful to this problem? $\endgroup$
    – Greg
    Commented Jun 1 at 5:19
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    $\begingroup$ Let $a \in G_1 \setminus \{1\}$, $b,c \in G_2 \setminus \{1\}$ with $b \ne c$. Then I think $aG_2a^{-1}$, $bG_1b^{-1}$, and $\langle cabac^{-1}ac^{-1} \rangle$ will work. Note that $bac^{-1}$ has infinite order, and hence so does $cabac^{-1}ac^{-1}$ even if $a^2=1$. It's just a matter of eliminating all possibilities for cancellation when you multiply elements. $\endgroup$
    – Derek Holt
    Commented Jun 1 at 12:32

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