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How many possible group of order 3 are there? I can only think of one such group.

David G. Stork
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Jaynot
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    Only one group of any prime order, because the order of a nonidentity element is a nontrivial divisor of the group's order. – Francis Begbie Jan 09 '16 at 23:38
  • @JohnMa I would say not exactly; the first answer on that question implicitly assumes Lagrange's theorem, and the writer admits that in the comments. Presumably the asker here hasn't yet seen/proven Lagrange's theorem. – Dustan Levenstein Jan 09 '16 at 23:43
  • Right, I guess if you didn't know already you should check that the cosets of a subgroup partition the group and all have the same size. – Francis Begbie Jan 09 '16 at 23:46
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    Try writing down the table of a group with elements $e$ (neutral element), $a$ and $b$. As you go step by step, you'll find that the group axioms force your hand, and there is only one way to fill the table. The $e$ column and row are easy. Then show that $ab$ can't possibly be $a$ or $b$. Continue that way for $ba$, $aa$ and $bb$. Remember that each element must appear exactly once in each row and in each column. – David Jan 09 '16 at 23:47
  • Related question: http://math.stackexchange.com/questions/228868/prove-that-a-group-of-order-3-must-be-cyclic – Olexandr Konovalov Jan 09 '16 at 23:57
  • My algebra professor used to call the groups of prime order "simple simple", because they have no (nontrivial) proper subgroups. If anyone feels that a solution of the earlier Question can be or should be given without explicitly relying on Lagrange's Thm., it could be added there. – hardmath Jan 10 '16 at 00:00
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    Jaynot, I think yours is a different question from the ones that have been described as duplicates. If you edit your question to say why those don't answer the question for you (e.g., they refer to "Lagrange's Theorem" and the definition of "cyclic" groups), then you can request that this question be reopened. – David Jan 10 '16 at 00:02

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