System of generators and surjective homomorphism

Let $G$ and $G'$ be groups and $\varphi:G\to G'$ a group homomorphism and $(g_s)_{s\in S}$ an indexed collection of elements of $G$ and is also system of generators of $G$. If $\varphi$ is surjective does it imply that $(\varphi(g_s))_{s\in S}$ is a system of generators of $G'$ ? Is there a simple example that the free group $F_n$ cannot be generated by fewer than $n$ elements. For example I cannot take $G$ and $G'$ cyclic and $\varphi$ such that it is not surjective, I must have more than $1$ generator, is it correct ?

• Yes. If $\varphi$ is surjective then image of the generators of $G$ generate $G^{\prime}$. I do not understand your sentence "Is there a simple example that the free group Fn cannot be generated by fewer than n elements." You cannot have an example of something which doesn't exist... – user1729 Nov 24 '14 at 10:18

The answer to the first part of the question is yes: since $\varphi$ is surjective for each $g' \in G'$ exists a $g \in G$ such that $$\varphi(g) = g'$$ since $(g_s)_{s \in S}$ are a family of generators for $G$ there's a finite sequence $s_1,\dots,s_n \in S$ and $i_1,\dots,i_n \in \{1,-1\}$ such that $g=g_{s_1}^{i_1}g_{s_2}^{i_2}\dots g_{s_n}^{i_n}$ and so $$g'=\varphi(g_{s_1}^{i_1}\dots g_{s_n}^{i_n})=\varphi(g_{s_1})^{i_1}\dots\varphi(g_{s_n})^{i_n}$$ hence the system $(\varphi(g_s))_{s \in S}$ is a system of generators for $G'$.
To address the second part you can consider the free group on $1$ generator, i.e. $\mathbb Z$, clearly $\mathbb Z$ cannot be generated by less then $1$ element, the only group with $0$ generators is the trivial group $(1)$ containing just the unit, which isn't isomorphic to $\mathbb Z$.
For another non trivial example consider the group $F_2$ this group cannot be generated by $1$ element, because otherwise it should be cyclic and so abelian, but it's a well known fact that $F_2$ is not commutative.
• Thanks a lot, I thought the same but group with $0$ generators was a bit confusing for me. – inequal Nov 24 '14 at 10:23
Let $\phi : G \to G'$ be a surjective homomorphism and let $E \subseteq G$ be a generating set. Then $\phi(E) \subseteq G'$ is also a generating set. You don't have to fiddle around with elements to see that (and the following argument also works in more general situations): If $H' \subseteq G'$ is a subgroup which contains $\phi(E)$, then $\phi^{-1}(H') \subseteq G$ is a subgroup which contains $E$. Since $E$ is a generating set, this means $\phi^{-1}(H') = G$ and hence $H' = \phi(\phi^{-1}(H'))=\phi(G)=G'$.