Bound for the cardinality of a minimal generating set of a finite group

Question

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)

Answer 1

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 $m\le\log_2(n)\le\sqrt{n}\le n/p$.