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Is there a systematic procedure for finding the smallest set of generating elements of a finite group?

Olexandr Konovalov
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lenzinho
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  • Consider that if the Group is of a prime order then every non unit element is a generating. Now try to think about the prime factors of N and how they help u. – JonesY Jul 09 '16 at 19:05
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    There is no general efficient procedure, and it can be difficult in some examples. But the number of generators of $G/[G,G]$ is a useful lower bound. Also, in many (but not all) groups, you will find a minimal set rapidly by repeatedly choosing random elements. – Derek Holt Jul 09 '16 at 19:06
  • Yes. Since the group is finite you could just check every set and find one that has the smallest size. I suspect this is not what you're looking for though, so if not you should try to be more specific. – Matt Samuel Jul 09 '16 at 19:06
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    @JonesY I am afraid that I have no idea what you have in mind! – Derek Holt Jul 09 '16 at 19:09
  • @DerekHolt : I think of using Sylow's Theorem of breaking the group into sub groups of $P^n$ order. – JonesY Jul 09 '16 at 19:16
  • @DerekHolt , (1) take random $log_2|G|$ but distinct element of $G$ , (2) find redundant element and refine until all non redundant element crate the the group. I am assuming the group $G$ is given. how is that? – Michael Jul 09 '16 at 19:47
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    @Jim That way you will find a generating set with no redundant elements. But it might not be a generating set of smallest possible cardinality. The group $A_5^{19}$ is a hard example. It is a $2$-generator group, but the proportion of pairs of elements that generate it is very small. – Derek Holt Jul 09 '16 at 20:19

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A following method can be used to find the smallest set of the generating elements in a finite group:

It is based on the following theorem:

Suppose $G$ is a finite group, $\{X_i\}_{i = 1}^{n}$ are i.i.d uniformly distributed random elements of $G$. Then $P(\langle \{X_i\}_{i = 1}^{n} \rangle = G) = \sum_{H \leq G} \mu(G, H) {\left(\frac{|H|}{|G|}\right)}^n$, where $\mu$ is the Moebius function for subgroup lattice of $G$.

Thus, the smallest possible cardinality of a generating set can be described by the following formula:

$$\min\{n \in \mathbb{N}| \sum_{H \leq G} \mu(G, H) {\left(\frac{|H|}{|G|}\right)}^n > 0\}$$

And if we know the smallest possible cardinality of a generating set (let's denote it by $s$), then we can find an example by checking all of the $C_{|G|}^s$ subsets on whether they lie in one of the maximal proper subgroups or not. If they lie - they are the smallest possible generating set.

Chain Markov
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  • Do you have a citation for the theorem? (Also, what does "i.i.d." stand for?) – user1729 Aug 26 '19 at 10:49
  • @user1729, "i.i.d." stands for "independent and identically distributed". The proof of the theorem can be found under the following link: https://math.stackexchange.com/a/2838078/407165 – Chain Markov Aug 26 '19 at 10:54