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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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marked as duplicate by Marc van Leeuwen, Paul, Ittay Weiss, Karolis Juodelė, Davide Giraudo Feb 5 '13 at 11:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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-Alvarez Feb 5 '13 at 8:14
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 5 '13 at 8:17
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 5 '13 at 8:23
See this for complete source as @Derek Holt showed. – S. Snape Feb 5 '13 at 8:27
up vote 2 down vote accepted

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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