The smallest finite group that can be generated by $n$ elements and cannot be generated by any less than $n$ elements is a product of $n$ cyclic groups of order $2$.

(a) Is there a largest finitely generated infinite group that can be generated by $n$ elements but not by more than $n$ elements? Largest in the sense that if you remove or change any relation between the generators you end up with a group that can be generated by more than $n$ elements.

ADDED: (a) is not really a question since any infinite group $G$ can be infinitely generated $<G>$. I'm stupid.

(b) Is there a smallest finitely generated infinite group generated by $n$ elements yet cannot be generated by less than $n$ elements? Smallest in the sense that any extra relation imposed on the group's generators will result in a finite group (or an infinite group generated by less than $n$ elements).

  • $\begingroup$ @PVAL see my edit. $\endgroup$ – user116457 Dec 24 '13 at 23:34
  • $\begingroup$ Isn't every finite Symmetric Group generated by 2 elements? A transposition and an n-cycle. Therefore, there does not exist a largest group generated by only 2 generators? $\endgroup$ – Josh B. Dec 25 '13 at 2:44
  • $\begingroup$ @JoshB. Thank you for making the question more precise. $\endgroup$ – user116457 Dec 25 '13 at 9:37
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    $\begingroup$ It seems the following. Suppose that we have a group $G$ generated by its subset $A$ and the set of relations $R$. If we add some extra relations $R_1$ to the set $R$ then the group $G’$ generated by the set $A$ with the family $R’=R\cup R_1$ of relations will be a quotient $G/N$ of the group $G$ with respect to its normal subgroup $N$. If $N$ is the unit subgroup, then all relations of the set $R_1$ are corollaries of the set $R$ of relations, so this case is trivial. $\endgroup$ – Alex Ravsky Dec 25 '13 at 15:47
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    $\begingroup$ But if $G$ is a simple group then each its normal non-unit subgroup $N$ coincides with $G$, so in this case $G’=G/G$ is the unit (that is a finite) group. $\endgroup$ – Alex Ravsky Dec 25 '13 at 15:47

Question (b) is equivalent to asking if there exists a group which is generated by $n$ elements (and no fewer) such that every non-trivial normal subgroup has finite index.

You can always add a group relation which is already implied by the group axioms or existing relations without changing your group. When you impose a non-trivial relation, you are quotienting by the normal closure of all of those elements which satisfy that relation.

There do exist such groups. The easiest example is $\Bbb Z$. All of its nontrivial quotients are finite, and it cannot be generated by 0 elements. Another example is the infinite dihedral group $D_\infty$ which requires 2 generators. In this group all normal subgroups have finite index, but there are still subgroups with infinite index. Also, any finitely-generated infinite simple group satisfies this requirement as mentioned in the comments. If we take a finite generating set, we can reduce it to a minimal generating set. We can then consider the collection of all minimal generating sets and then set $n$ to be the minimum of those cardinalities.

If you only want to consider quotients which have fewer generators than the original group, that seems to be a difficult question. In finitely-generated non-abelian groups, you can have minimal generating sets of different cardinality. And you can even have subgroups which require more generators than the whole group itself does.


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