0
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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... $\endgroup$ – user1729 Nov 24 '14 at 10:18
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks a lot, I thought the same but group with $0$ generators was a bit confusing for me. $\endgroup$ – inequal Nov 24 '14 at 10:23
  • 1
    $\begingroup$ You forgot inverses of the generators. $\endgroup$ – Martin Brandenburg Nov 24 '14 at 10:51
  • $\begingroup$ @MartinBrandenburg ops... thanks for pointing out, I'm going to edit immediately. $\endgroup$ – Giorgio Mossa Nov 24 '14 at 11:13
2
$\begingroup$

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 $H' = \phi(\phi^{-1}(H'))=\phi(G)=G'$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.