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Consider $G=F \rtimes T$, where $F=\mathbb{Z}_3 \times \mathbb{Z}_3$ and $T=\mathbb{Z}_5$. Let $\phi : \mathbb{Z}_5 \rightarrow Aut(\mathbb{Z}_3 \times \mathbb{Z}_3)$.

It is said that any group is the quotient of a free group? How can I represent the above group $G$ as a quotient of a free group? Do I have to specify $\phi$?

If I know $\phi$ can some one explain what steps I should follow to represent it as a quotient of a free group?

An answer to an example of $\phi$ is also Ok.

Many thanks in advance.

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    $\begingroup$ You speak about $G$ but you didn't specify $\phi$, which determines the semidirect product. $\endgroup$
    – J. De Ro
    May 30, 2020 at 11:38
  • $\begingroup$ Yes, @ε-δ can you suggest an example for $\phi$ and explain answer to that? I have been studying about the semidirect product only for a very short period of time and that I have studied from a visual group theory book. $\endgroup$
    – Bob Traver
    May 30, 2020 at 15:46
  • $\begingroup$ There we can arrange the vertices of a Cayley graph such that the vertices corresponding to the normal subgroup and its cosets can be arranged such that copies of the subgraph including the vertices of the normal subgroup can be arranged along a $5$-cycle. But this visual idea is not relevant to the above question anyway.. $\endgroup$
    – Bob Traver
    May 30, 2020 at 15:46

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You have to specify $\phi$ when there are different groups that are not isomorphic between them.
In this case $Aut(\mathbb{Z}_3 \times \mathbb{Z}_3) \cong GL(2,3)$ but the order of $GL(2,3)$ is $(3^2-1)(3^2-3)=48$ so the only $\phi$ is the trivial one.
You can always represent a finite group as a quotient of a free group; you have to work with the generators of the group.
In this case your group is isomorphic to $\mathbb{Z}_5 \times \mathbb{Z}_3 \times \mathbb{Z}_3$ so there are $3$ generators, namely $a$,$b$,$c$.
Since the group is abelian, the subgroup of $F_3$ you quotient for containes the derived group of F, so it containes $\{ [a,b] ; [a,c] ; [b,c] \}$.
Morover we have to impose the condition about the order of the generators, so the subgroup have to contain $a^5$ , $b^3$ and $c^3$.
So, you can represent this group as $$ \frac{F_3}{\langle a^5,b^3,c^3,a^{-1}b^{-1}ab,a^{-1}c^{-1}ac,b^{-1}c^{-1}bc \rangle }$$

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  • $\begingroup$ Thank you very much @NotPhiQuadro can you please kindly write an example for a case where $F$ and $T$ are interchanged? i.e. $T=\mathbb{Z}_3 \times \mathbb{Z}_3$ and $F=\mathbb{Z}_5$. Because the above is trivial case, but I would like to try to semidirect product? $\endgroup$
    – Bob Traver
    May 31, 2020 at 20:13
  • $\begingroup$ Can you please mention a specific $\phi$ in your example and show how to get the free group? Many many thanks in advance. $\endgroup$
    – Bob Traver
    May 31, 2020 at 20:14
  • $\begingroup$ If you switch $T$ and $F$ the problem is excatly the same because the order of $Aut(\mathbb{Z}_5)$ is $4$ and the only function $\phi$ from $T$ to $Aut(\mathbb{Z}_5)$ is the trivial one, so, also in this case, the only semidirect product you can choose is the direct product. $\endgroup$
    – Sant97
    May 31, 2020 at 21:51
  • $\begingroup$ You can try different groups, but generally is a non trivial task finding the representation of this groups if you don't know more things about them $\endgroup$
    – Sant97
    May 31, 2020 at 21:52

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