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Finding a minimal generating set of a finite group is difficult but in the case of abelian groups using the fundamental theorem of finite abelian groups, we can find it easily. Moreover, corresponding to each $k\in \mathbb N$, we have an abelian group such that the cardinality of a minimal generating set is $k$ (take rank $k$ abelian groups). In the case of non-abelian groups, we know that symmetric group on $n$-letters $S_{n}$ can be generated by two elements. Can you give an example of a finite non-abelian group that cannot be generated by two elements? In general, can we find a nonabelian group corresponding to each $k>1$ such that the cardinality of a minimal generating set is $k$?

Thank you.

MathMeet
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  • Look up a wiki about the Burnside Problem and Burnside Groups. For example https://encyclopediaofmath.org/wiki/Burnside_group – Nicky Hekster Mar 22 '22 at 09:00
  • Or see here for examples https://mathoverflow.net/questions/380840/can-a-direct-product-of-nonabelian-simple-groups-be-generated-by-two-elements – Nicky Hekster Mar 22 '22 at 09:06
  • Perhaps the simplest examples are the extraspecial groups of order $p^{2n+1}$, where $p$ is a prime, all of whose minimal generating sets have order $2 n$. A central product of such a group by a cyclic group of order $p^{2}$ allows you to get odd numbers of generators as well. – Andreas Caranti Mar 22 '22 at 09:49
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    "Minimal generating set" is ambiguous. Minimal with respect to inclusion or cardinality.? – Derek Holt Mar 22 '22 at 10:51
  • @AndreasCaranti Thank you for the answer. One more example I am thinking of: What can we say about a minimal generating set of direct products of an abelian group of rank $k$ and a finite non-abelian group(like any symmetric group)? Clearly, the group is non-abelian. Can we comment on its minimal generating set now? – MathMeet Mar 22 '22 at 10:57
  • @DerekHolt I am using this definition. https://groupprops.subwiki.org/wiki/Minimal_generating_set#:~:text=A%20generating%20set%20of%20a,dropped%20from%20the%20generating%20set. – MathMeet Mar 22 '22 at 11:00
  • Isn't $S_n\times (\mathbb{Z}_p)^k$ (for some prime $p$ not dividing $n!$) generated by $k+2$ elements and not less? And yet nonabelian. – freakish Mar 22 '22 at 11:33
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    @Kavita That definition is minimal with respect to inclusion. That makes your question easier because, for example, $S_n$ has minimal sets of different cardinalities, in fact all $k$ with $2 \le k \le n-1$. – Derek Holt Mar 22 '22 at 11:44
  • @freakish I am thinking the same. Not sure! How to prove it if it is exactly $k+2$ not less? – MathMeet Mar 22 '22 at 12:10
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    @Kavita Not sure how to prove what? AndreasCaranti has explained how to do this in his comment. – Derek Holt Mar 22 '22 at 14:36
  • @Kavita, if $A$ is a finite abelian group with minimal number of generators $k$, and $S$ is a non-abelian group, then a minimal generating set of $G = A \times S$ has at least $k$ elements, as such a set projects onto a set of generators of $G/S \cong A$. However, if $A$ is elementary abelian of order $2^6$ (so $k = 6$ here), one can check that when $S = A_{4}$ then the minimal number of generators of $G$ is $6$, and when $S = S_{4}$ the number is $7$. – Andreas Caranti Mar 22 '22 at 15:46

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You are using the following definition:

A minimal generating set for a group $G$ is a set $S$ such that $\langle S\rangle = G$, but for every proper subset $T$ of $S$, $\langle T\rangle\neq G$.

As Derek Holt points in comments, this makes an affirmative answer to your question easy: the set $$\{(1,2), (1,3),\ldots,(1,k-1), (1,k), (1,k+1)\}$$ is a minimal generating set with $k$ elements for $S_{k+1}$.

In fact, $S_n$ has minimal generating sets of sizes $k$, $2\leq k\leq n-1$: the set $\{(1,2),(1,3),(1,4),\ldots,(1,k),(k,k+1,\ldots,n)\}$ is easily verified to a minimal generating set.

Moreover, as noted in comments here, it is a theorem that if a finite group $G$ has a minimal generating set with $k$ elements, and a minimal generating set with $\ell\gt k$ elements, then it has minimal generating sets with $m$ elements for all $m$, $k\leq m\leq\ell$. This is Tarski's Irredundant Basis Theorem. See references cited in this MathOverflow question about the values of $D(G)$, the size of the largest possible minimal generating set for $G$.

On the other hand, given a finite group $G$, if we let $d(G)$ denote the cardinality of the smallest minimal generating set for $G$ (also called the "rank of $G$" or the "generating rank of $G$"), you might ask whether for any $k>1$ there exists a finite nonabelian group such that $d(G)=k$. Again, as noted in the comments, an extra-special $p$-group ($p\gt 2$) of order $p^{2n+1}$ has generating rank $2n$, and then taking the direct product with a cyclic group of order $p$ gives a group with generating rank $2n+1$, so the answer is "yes" here as well.

Arturo Magidin
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