Abelian group is not cyclic Suppose that $n\ge3$ where n is odd and $D_n$ is a dihedral group of order $2n$. I need to prove or disprove that if $H$ is an abelian finite subgroup of $D_n$, then $H$ is cyclic. I cannot understand what does $H$ being a finite subgroup of dihedral group means. Like, if $D_3=[e,r,r^2,s,sr,sr^2]$, then how can I construct an abelian subgroup of this so that it is not cyclic? Can someone elaborate on this please?

Thanks in advance!
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.
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.