Let us recall some definitions:
One says that $H$ is a subgroup of $D_n$ (or any group) when $H$ is a subset of $D_n$ this is itself a group with respect to the inherited operations. More explicitly it is a subset that contains the identity element and is closed under products and taking inverses. That it is finite just means that the set $H$ is finite; in this case this is automatic as every subset of $D_n$ is finite.
A group (or subgroup) is called abelian when the group law satisfies the claw of commutativity.
Thus what you are supposed to decide is: if $H$ is a (finite) subset of $D_n$ that contains the identity and is closed under multiplication and taking inverses (that is, a subgroup) and such that $gh = hg$ for all $h,g \in H$ (so that it is an abelain subgroup), is it then true that $H$ is cyclic, that is it is can be generated by a single element.
It is true that to disprove this it would suffice to exhibit an example of an abelian subgroup of $D_n$ that is not cyclic.