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ndkrempel
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Although it doesn't directly answer your question, it seems interesting to consider groups in which every normal generating set is already a generating set.

At least in the finitely generated world, this is equivalent to "every maximal subgroup is normal", which is also equivalent to $G' \leq \Phi(G)$. For finite groups, this is equivalent to being nilpotent, but for finitely generated groups it may be a strictly weaker condition than nilpotence (I don't know an example however.)

In any case, such a group (for example, any f.g. nilpotent group) necessarily has $\mathrm{nr}(G) = \mathrm{rank}(G)$.

Although it doesn't directly answer your question, it seems interesting to consider groups in which every normal generating set is already a generating set.

At least in the finitely generated world, this is equivalent to "every maximal subgroup is normal", which is also equivalent to $G' \leq \Phi(G)$. For finite groups, this is equivalent to being nilpotent.

In any case, such a group necessarily has $\mathrm{nr}(G) = \mathrm{rank}(G)$.

Although it doesn't directly answer your question, it seems interesting to consider groups in which every normal generating set is already a generating set.

At least in the finitely generated world, this is equivalent to "every maximal subgroup is normal", which is also equivalent to $G' \leq \Phi(G)$. For finite groups, this is equivalent to being nilpotent, but for finitely generated groups it may be a strictly weaker condition than nilpotence (I don't know an example however.)

In any case, such a group (for example, any f.g. nilpotent group) necessarily has $\mathrm{nr}(G) = \mathrm{rank}(G)$.

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ndkrempel
  • 1.8k
  • 15
  • 23

Although it doesn't directly answer your question, it seems interesting to consider groups in which every normal generating set is already a generating set.

At least in the finitely generated world, this is equivalent to "every maximal subgroup is normal", which is also equivalent to $G' \leq \Phi(G)$. For finite groups, this is equivalent to being nilpotent.

In any case, such a group necessarily has $\mathrm{nr}(G) = \mathrm{rank}(G)$.