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Trying to find a contradiction with out writing out a cayley table. So far, I understand that the group would be commutative. However I believe I fail to understand how closure is broken or in other words how there are more elements generated than their should be.

I do understand that there are suppose to be two groups of order 6. The S3 group and the cyclic group. But my primary motivation is to understand why this does not work.

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    $\begingroup$ Such a problem can be approached in many ways, but you've given little for the Reader to know what you have already tried, what your motivation or interest in the problem is, etc. Please make the question body as self-contained as possible (not relying on the title to carry the burden of posing a problem) and add more context so that responses can address some specific difficulty you have. $\endgroup$ – hardmath May 2 '15 at 15:00
  • $\begingroup$ Thanks you! I just edited it to be more clear. Answers below have been helpful. $\endgroup$ – grilledsardines May 2 '15 at 18:20
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Pick two elements of order two. Their product is also an element of order two. So they generate a subgroup of order 4 isomorphic to the Klein 4 group. But a group of order 6 cannot have a subgroup of order 4 as 4 does not divide 6.

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If you know Cauchy's theorem then you know that every finite group of order $n$ has an element of order $p$, where $p$ is any prime dividing $n$.

In this case this theorem forces the existence of an element of order $3$.

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@Gamamal 's answer is correct, and general. You could answer the question in your special case by checking it for the only two groups of order 6. One is cyclic, the other is the set of permutations of three elements.

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