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It seems to be known that the minimum size of a generating set for a finite group of order $n$ is at most $\log_2 n$. Can someone explain why this is true?

Edit: noted that the logarithm is base 2, and $n$ represents the order of the group.

argentpepper
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  • You mean: the maximum over all groups of order $\le n$ of the minimal size of a generating subset. – YCor Nov 03 '12 at 20:22

2 Answers2

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If $G$ is a finite group, and $\{g_1,\ldots,g_m\}$ is a minimal set of generators, let $G_n = \langle g_1,\ldots,g_n \rangle$. Then by minimality $g_k \ne e$ for all $k$, and $g_{n+1} \notin G_n$ for all $n$. Then $G_{n+1}$ contains at least the two disjoint cosets $eG_n=G_n$ and $g_{n+1}G_n$ of $G_n$, so $|G_{n+1}| \ge 2|G_n|$. By induction $|G| = |G_m| \ge 2^m$, so $m \le \log_2 |G|$. This inequality is sharp for $G = \mathbb{Z}_2^m$.

Lukas Geyer
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  • Why must there be an element $g_{n + 1}$ not in $G_n$?

    Also, can you explain why $|G_{n + 1}| \geq 2 |G_n|$? Why must the intersection of $eG_n$ and $g_{n + 1}G_n$ be empty?

     – argentpepper Nov 27 '12 at 19:33
  • If $g_{n+1} \in G_n$, then you could remove $g_{n+1}$ from the set of generators and still generate the whole group, so this would violate minimality of the set of generators. The fact that the cosets are disjoint is a general fact in group theory. If $g \in G_n \cap g_{n+1} G_n$, then $g = g_{n+1} h$ with $h \in G_n$, so $g_{n+1} = g h^{-1} \in G_{n}$, a contradiction. – Lukas Geyer Nov 27 '12 at 23:01
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    By the way, this simple and elegant argument is given in this MathSE answer of Feb. 2012 by Geoff Robinson, who attributes it to P.M. Neumann. It is used there to bound the size of the automorphism group (it implies that the size of the endomorphism monoid is $\le |G|^{\log_2|G|}=2^{\log_2(|G|)^2}$). – YCor Nov 04 '24 at 15:17
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Think about the smallest group which needs $r$ generators - you will find that each generator has order 2, and that the group has order $2^r$ (this isn't a proof, but an indication of a method). Your statement is only true, therefore, if your logarithms are being taken to base 2.

Mark Bennet
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