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Smallest possible odd integer that can be the order of a non-Abelian group.

Attempt: A non abelian group means $Z(G) \subset G$ . Hence, it suffices to find the smallest odd integer $n$ such that $Z(G)$ is a proper sub group of $G$.

Unfortunately, I do not have a strategy in mind beyond this. How should I move ahead?

Thank you for your help.

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Hint: Recall that groups of order $pq$ and $p^2$ for $p,q$ primes with $p<q$ and $q\not\equiv 1 \pmod{p}$ are abelian. This rules out many "small" numbers.

Now think about the small odd-order non-abelian groups that you know of -- can you rule out all possible orders less than that one of these examples, using the first part?

Alternately, if you're having trouble coming up with appropriate examples, pick the minimal order that you haven't ruled out. Try to explicitly construct a non-abelian group of that order (or show that one cannot exist for other reasons).

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    $\begingroup$ Thank you for the answer :-) $\endgroup$ – MathMan Aug 26 '14 at 18:51

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