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How to show that $\mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/3\mathbb{Z}$ contains a free group on two generators.

I do not how to approach this problem, How to pick the generators of the free group ?

Edit (following comments): We know free product of cyclic group of order $2$ and cyclic group of order $3$ i.e. $\mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/3\mathbb{Z}$ is isomorphic to $PSL(2,\mathbb{Z})$. The generators of that free group are \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} and \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}

Now I know $PSL(2,\mathbb{Z})=SL(2,\mathbb{Z})/\{+I, -I\}$.

Tensor_Product
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  • Aren't free groups infinite in size? You should explain more what your thought is. – hardmath Oct 02 '16 at 20:50
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    @hardmath This product is the set of all words written with letters in the two groups, not ordered pairs of letters. It will be infinite. – Aaron Oct 02 '16 at 20:52
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    (Following your edit) If you have an isomorphism between your group and $PSL(2,\mathbb Z)$, why not just take the image of those two matrices under the isomorphism? Your edit reduces the problem to a previously solved problem, which is the way of the mathematician! – Aaron Oct 02 '16 at 21:03
  • @Aaron But I have only shown that, $SL(2,\mathbb{Z})$ contains a free group of rank $2$ not $PSL(2,\mathbb{Z})$. Is it true that if I consider the equivalence class of the generators in $PSL(2,\mathbb{Z})$, It generate a free group in $PSL(2,\mathbb{Z})$. – Tensor_Product Oct 02 '16 at 21:06
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    ${\rm PSL}(2,{\mathbb Z})$ is a quotient of ${\rm SL}(2,{\mathbb Z})$ by a subgroup of order $2$. Since free groups are torsion-free, if ${\rm SL}(2,{\mathbb Z})$ contains a free subgroup then so does ${\rm PSL}(2,{\mathbb Z})$. – Derek Holt Oct 02 '16 at 21:14
  • @DerekHolt If you expand and post this as answer (which the site guidelines explicitly ask for since it is not a comment, it is an answer), then https://math.stackexchange.com/q/5035551/1650 can be marked as a duplicate. – Martin Brandenburg Feb 15 '25 at 12:53

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The kernel of the projection $\rho: {\rm SL}(2,{\mathbb Z}) \to {\rm PSL}(2,{\mathbb Z})$ is the subgroup $\{\pm I_2\}$ of order $2$.

Let $F < {\rm SL}(2,{\mathbb Z})$ be isomorphic to the free group of rank $2$. Since $F$ is torsion-free, $F$ has trivial intersection with $\{\pm I_2\}$, and so $\rho(F) \cong F$ is a free subgroup of rank $2$ in ${\rm PSL}(2,{\mathbb Z})$.

Derek Holt
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