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Three finite groups with the same numbers of elements of each order

With any finite group $G$ I can associate a multiset $S_G = \{\text{ord}(g) : g \in G\}$. Is the map $G \mapsto S_G$ injective? Is there an algorithm to generate a multiplication table of $G$ from $S_G$? Given any multiset $S$, how hard is it to check if it represents some group in this way?

Having tried a few groups, it does look like an injection, but a formal proof or a construction (brute force aside) escapes me.

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Karolis Juodelė
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    See http://mathoverflow.net/questions/39848/finite-groups-with-elements-of-the-same-order – Mariano Suárez-Álvarez Feb 05 '13 at 08:12
  • In short, there do exist both finite non-abelian groups and abelian groups of the same order in which all non-trivial elements have order $p$, for some primes $p$. – Mariano Suárez-Álvarez Feb 05 '13 at 08:14
  • This is clearly false, as other have pointed out- it fails badly for $p$-groups for odd $p.$ I believe it is an open question if attention is restricted to finite simple groups. – Geoff Robinson Feb 05 '13 at 08:17
  • Well, that's a pity. My last questions remains somewhat relevant though. (Given any multiset S, how hard is it to check if it represents some group in this way?). – Karolis Juodelė Feb 05 '13 at 08:23
  • See this for complete source as @Derek Holt showed. http://math.stackexchange.com/q/159395/8581 – Mikasa Feb 05 '13 at 08:27
  • This question is an exact duplicate of http://math.stackexchange.com/questions/181632/, which has been closed as a duplicate of http://math.stackexchange.com/questions/62331/three-finite-groups-with-the-same-numbers-of-elements-of-each-order (although that really is a different question). Since the system will not allow me to vote to close this questoin as a duplicate of the first one, I'll vote to close it as a duplicate of the second one. By the way I found this because I was looking for a much more recent duplicate that I saw, but searching is so miserable that I cannot find it. – Marc van Leeuwen Feb 05 '13 at 10:11

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For a simple counterexample, $\mathbb{Z}_4\times \mathbb{Z}_4$, $\mathbb{Z}_4\rtimes \mathbb{Z}_4$, and $Q_8\times \mathbb{Z}_2$ all have the same multisets of element orders. In general, the list of element orders does not tell you very much about a group's structure.

I addressed your second question here and your third question here.

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Alexander Gruber
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