Converse of Theorem Proof/Counterexample

Question

Theorem: If $G$ is an abelian group such that $H\leq G$, then $H$ is an abelian subgroup.

I've proved this theorem here:

Proof: Suppose $G$ is abelian. Let $a,b\in G$. Then $ab=ba$ as G is abelian and commutative properties are inherited. Since $H\leq G$, then for all $a,b\in H, ab=ba$ also. Thus, $H$ is abelian.

I am new and still learning how to construct proofs -- please be kind!

The question: Is the converse of the theorem true?

I am having a hard time understanding exactly what the text means by just the converse of the theorem. Are they asking "If $G$ is not an abelian group such that $H\leq G$, then $H$ is not an abelian subgroup? Or, are they asking, "If $G$ is an abelian group such that $H\leq G$, then $H$ is not an abelian subgroup?

If $G$ is an abelian group and if $H\leq G$, then can $H$ be an abelian subgroup? Well, isn't this kind of trivial. Suppose $G$ is abelian. Then laws of commutative properties and other group properties follow; that is, there exist an inverse, identity, and $H$ would also be closed under the binary operations *, where the operation * is associative. Then, wouldn't all elements of $H$ also be abelian since they inherit all these properties from $G$?

Answer 1

The converse of a theorem is the statement you obtain when you interchange the hypothesis and the thesis. In this case, the converse should be: "If $G$ is a group such that $H \leq G$ is an abelian subgroup, then $G$ is abelian". You should easily be able to find a counterexample for this claim.

Answer 2

The converse, $G$ has an abelian subgroup implies $G$ is abelian, is false. Consider $V_4\le S_4$ (either the normal one, or one of the nonnormal ones). Or, any cyclic subgroup of any $S_n$. Or any cyclic subgroup of your favorite nonabelian group.

The other possible interpretations appear to be trivial.