Maximum number of minimal set of generators of groups of order $n$

Question

I was discussing with my friend this curious problem: Consider the number of generating elements for groups with order $n$. We define $R_n$ as follows:

$$ R_n = \{k \in \mathbb{Z}^+ \mid\exists\text{ a group $G$ of order $n$ such that $G$ is generated by $k$ elements minimally}\} $$

Here, I define "generated by $k$ elements minimally" as in the minimal number of elements required to generate $G$.

It is obvious that $\inf{R_n} = 1$ always, as we can take the cyclic group. What about $\sup R_n$? What is the behaviour of the $\sup R_n$ as $n \to \infty$? Perhaps there is a "nice" function $f$ such that $\frac{\sup{R_n}}{f(n)} \to c$ for some constant $c$?

Any input would be appreciated!

Answer 1

It tends to infinity: $\sup R_n\rightarrow\infty$ as $n\rightarrow \infty$.

One way of seeing this is to consider direct products. For a group $G$, write $d(G)$ for the minimum number of generators needed to generate $G$, and write $G^ k$ for the direct product of $k$ copies of $G$. The following is a theorem of Weigold*, but is very close to a result of Dey**.

Theorem 2.1. For every non-trivial finite group $G$ and every positive integer $k$, $d(G^k) > \log_{|G|}k$.

So we have that $R_{|G|^k}\geq d(G^k) > \log_{|G|}k$. As $\log_{|G|}k\rightarrow\infty$ as $k\rightarrow\infty$, the result follows.

*Theorem 2.1 of Wiegold, James. "Growth sequences of finite groups." Journal of the Australian Mathematical Society 17, no. 2 (1974): 133-141. link

**Dey, I. M. S. "Embeddings in Non‐Hopf Groups." Journal of the London Mathematical Society 2, no. 1 (1969): 745-749.

Answer 2

Hint $(\mathbb Z/ p \mathbb Z)^n$ is a vector space of dimension $n$ over $(\mathbb Z/ p \mathbb Z)$. Therefore, it needs at least $n$ generators.

Answer 3

It's easy to show that a group of order $n$ can be generated with $\lfloor \log_2(n)\rfloor$ elements. (This follows from Lagrange's Theorem.) The groups $C_2^m$ show that this bound is tight. This question has been asked many times on this site, and the answer is easy to find by googling.