# Non-Abelian groups with all proper subgroups Abelian

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.

• Well, try it with $S_3$ for example. – lulu Mar 27 '18 at 18:44
• 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 $H \neq G$. – Student Mar 27 '18 at 18:47
• 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. – YCor Mar 28 '18 at 12:22

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.
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.