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I need to show that if $G$ is non-abelian group of order $6$ then it contains an element of order $3$. I don't know how to proceed. Any kind of help/hint is appreciated.

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user311359
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    there are only two order six groups: $\Bbb Z_6$ and $S_3$ – janmarqz Feb 07 '16 at 02:21
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    Hint: suppose not. Then every element (other than the identity) has order $2$. Show that such a group is always abelian. – lulu Feb 07 '16 at 02:23
  • Can't you just use Cauchy's theorem? – Noble Mushtak Feb 07 '16 at 02:26
  • Why G is given to be non-abelian? What goes wrong in case of G being abelian? – user311359 Feb 07 '16 at 02:28
  • The Sylow theorems tell you that there is a subgroup of order 3. As 3 is prime, each element which is not the neutral one in this subgroup is a generator of it. – johnnycrab Feb 07 '16 at 02:39
  • @user311359 Do you know results like the Sylow Theorems? If you do, then your question certainly follows quickly from them. But your special case is a lot easier than the general theorem. In my hint, I sketched one way to do it...I'm sure there are others. – lulu Feb 07 '16 at 02:44

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As noted in the comments, with the Sylow theorems the solution is trivial. So I would go with lulu's hint: suppose not and show that such a group is always Abelian. Then every non-neutral element has order $2$. The key then is to be aware that for such an element $a$, $a^2 = e \implies a^{-1}=a$.
After that, a little fooling around with your elements will quickly solve your problem.

$$\forall a,b \in G: (ab)(ab)^{-1} = e \implies abab = e \implies aba = b \implies ab = ba$$

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johnnycrab
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