TITLE: Do orders uniquely describe a group?
QUESTION [7 upvotes]: Possible Duplicate:
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.

REPLY [3 votes]: 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.