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Let $G$ be a finite group. It can be written as an unique direct sum of indecomposable groups $G_1 \times \cdots \times G_m$

Is there a method to find these groups $G_1\cdots G_m$ if I can only :

  • pick a random $G$ element $g$
  • given $g$, get $-g$
  • given $g_1,g_2, $ get $g_1 + g_2$

Thank you for your help

Cyrius Nugier
  • 69
  • Is this an abelian group? If not, why are using additive notation? There cannot be a deterministic method, because your random element generator might keep returning the identity element! In these types of problems, it is standard to assume that you are given a generating set for the group. – Derek Holt Jun 22 '20 at 14:29
  • what kind of operations are prohibited? – miracle173 Jun 22 '20 at 14:32
  • It was not supposed to be an Abelian group, what is the correct notation ? If you are given generators, does this become really easier ? – Cyrius Nugier Jun 22 '20 at 14:42
  • Since the group is finite, there is a nonzero probability of picking the same (identity) element all the time. So I think that the problem is hopeless. – markvs Jun 22 '20 at 14:59
  • And if it gives a new $G$ element every time ? – Cyrius Nugier Jun 22 '20 at 15:06
  • If it gives a new element every time, repeat until you get the same element twice. You then have $G$ as a set, and the other two operations allow you to construct the group table. – rogerl Jun 22 '20 at 16:44
  • If it gives a new element each time then it is not generating random group elements. It is very unclear what you really want to ask and why. What you are describing sounds very like a black-box group as defined by Babai, but you have to assume that you are given a generating set for $G$. – Derek Holt Jun 22 '20 at 17:12

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