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.
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 $$ \frac{F_3}{\langle a^5,b^3,c^3,a^{-1}b^{-1}ab,a^{-1}c^{-1}ac,b^{-1}c^{-1}bc \rangle }$$