Let $f(n)$ denote the order of the smallest finite group which cannot be generated with less than $n$ elements. Trivially $f(n) \leq 2^n$ since ${\mathbb F}_2^n$ can be seen as a vector space with dimension $n$. Is the exact value of $f(n)$ known?
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asked Nov 2, 2011 at 13:10
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$\begingroup$ In general a group $G$ of order $n$ can be generated by at most $\log_2(n)$ elements; so if $n< 2^k$, then this minimal set has cardinality smaller than $k$. $\endgroup$– user641Nov 2, 2011 at 21:32
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1 Answer
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$f(n) = 2^n$
Let $X$ be the set of n generators of the finite group $G$ where $G$ cannot be generated by $n-1$ elements. Consider a sequence of subgroups of $G$, the first subgroup being $G$ itself generated by all of $X$, each of the next subgroups generated by one fewer generator than the last, finally down the trivial group. Each subgroup in the sequence must have distinct order which divides the order of the previous subgroup in the sequence. Thus the order of $G$ must be the product of $n$ terms all greater than one. Thus the smallest possible order of $G$ is $2^n$.
