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Given a finite group $G$ of order $n\geq 2$, it obviously has a minimal generating set (a generating set of minimal cardinality) let's say of cardinalty $m$. I am looking to find out if a bound for $m$ can be established. I asked myself this question after I saw a problem which asked to prove that $$|\text{End}(G)|\leq\sqrt[p]{n^n}$$ Where End$(G)$ is the set of endomorphisms of $G$ and $p$ is the greatest prime divisor of $n$. Given the fact that an endomorphism would be uniquely determined by its values on the elements of a generating set of $G$ that problem would be solved by showing that $$m\leq \frac{n}{p}$$ which is an idea for a possible bound. (I'm not looking to find another solution to the problem above, I only mentioned it because it provided an idea for a bound)

John WK
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  • Is $m\le n-1$ good enough for you? – Somos Apr 02 '19 at 20:18
  • Related: https://mathoverflow.net/questions/76417/size-of-smallest-generating-set-of-a-group – Arturo Magidin Apr 02 '19 at 20:20
  • We always have $m \le \log_2 n$. I will leave the proof (which is quick and elementary) as an exercise (but I am sure this has been asked before). – Derek Holt Apr 02 '19 at 21:18
  • "minimal generating subset" does not mean generating subset of minimal cardinal; it's weaker and means minimal for inclusion among generating subsets. For instance ${2,3}$ is a minimal generating subset of $Z/6Z$. Derek's inequality holds for $m$ being the cardinal of any minimal generating subset. – YCor Apr 02 '19 at 22:44
  • @YCor In my lengthy experience, the convention in mathematical circles is that anyone can define anything to mean anything they like, and since the poster has at least taken the trouble to define "minimal generating set", one cannot complain. An example of something that annoys me along these lines is "strictly less than", because it raises the question of what "non strictly less than" might mean! – Derek Holt Apr 02 '19 at 22:54
  • If there's consensus on the meaning of a definition and I read a question using this definition, it's likely that I'll not pay attention to its definition even if recalled, at least if the question looks coherent with the definition I expect. – YCor Apr 02 '19 at 22:59

1 Answers1

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Let $m$ be the cardinal of a minimal generating subset in a group of order $n\ge 2$ and $p$ the minimal prime divisor of $n$.

If $n$ is prime, then $m=1=n/p$.

If $n$ is not prime, then $n/p\ge\sqrt{n}$ and also $n\ge 4$, which implies $\log_2(n)\le\sqrt{n}$. Combined with the inequality mentioned by Derek Holt, you deduce $m\le\log_2(n)\le\sqrt{n}\le n/p$.

YCor
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  • PS: the inequality mentioned by Derek also more directly gives the better inequality $|\mathrm{End}(G)|\le n^{\log_2 n}$ for $|G|=n$, which is attained for $n$ a power of 2, for the 2-elementary group $G$ of order $n$. – YCor Apr 03 '19 at 08:45