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Given any non abelian group, how can I prove that every proper subgroup may be abelian? I know the definition of "abelian," but I don't know the difference between a group and a subgroup, nor do I understand how the two interconnect.

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    $\begingroup$ Well, try it with $S_3$ for example. $\endgroup$ – lulu Mar 27 '18 at 18:44
  • $\begingroup$ Let $G, \ast$ be a group and $H$ a subset of $G$, then $H$ is a subgroup of $G$ if $H, \ast$ forms a group. An equivalent statement is that $H$ is not empty and for all $x,y \in H$ we have $x \ast y^{-1} \in H$. A subgroup $H, \ast$ of a group $G, \ast$ is proper if $H \neq G$. $\endgroup$ – Student Mar 27 '18 at 18:47
  • $\begingroup$ A subgroup of a group $G$ is a particular case of a group. It is a group that occurs as subset of the given group $G$, with the induced law. It's a very practical notion, because it allows to avoid checking axioms such as associativity. $\endgroup$ – YCor Mar 28 '18 at 12:22
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Hint If $H$ is a proper subgroup of $G$ then $|H|$ is a proper divisor of $|G|$.

Hint 2 If all the proper divisors of $|G|$ are prime, then all the proper subgroups of $G$ are cyclic.

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There are many non-abelian groups all of whose proper subgroups are abelian. Studying such groups of low order, we immediately find examples, such as $S_3$ or $Q_8$, the quaternion group. Because we know all subgroups explicitly for these groups, it is easy to prove that they are abelian.

One might ask what properties this class of groups has. Every such group is necessarily metabelian, for example.

More generally there is the following reference here:

What can we say of a group all of whose proper subgroups are abelian?

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