Abelian Subgroup and proposition $ab^{-1}$ 
If $G$ is an abelian group and $n\in\mathbb{N}$, show that $H=\{x\in G:x^n=e\}$ is a subgroup of $G$.($e$ stands for  the neutral element. )

First I check if H  is a subgroup, by applying the following theorem:

Proposition: Let $G$ be a group and $H$ a non-empty subset of $G$. $H$ is a subgroup of $G$ iff $\forall a,b\in H, ab^{-1}\in H$

Proof:
1) As $H\neq \emptyset$, there is a $x\in H$ such that $xx^{-1}\in H$, since $a,b$ are defined as arbitrary. 
2) Since $x\in G$ and $G$ is a group then $xx^{-1}=e$, so $e\in H$. 
Now I prove that $H$ is closed under the operation defined in G. 
3) If $x,y\in H $ so it is $xy^{-1}$, by 2), it follows $x(y^{-1})^{-1}=xy\in H$.
Associativity follows by the fact $H$ is a subset of $G$.
Proof problem: Now going back to the problem:
Since $H\subset \{\forall x\in G\}$, the H is abelian. If $x,y\in H$, $x^ny^{-n}\in H$ then H is a subgroup of G. $x^ny^{-n}=e(y^{n})^{-1}=e$, proving that $H\leq G$. 
Questions:
a) I would like if someone could check my proof specially the proposition proof.
b) What do you think about a subgroup of an abelian group being necessarily an abelian subgroup? Is this true?
Thanks in advance!
 A: In the proof of the proposition, I understand that 1) and 2) together are meant to show that $e\in H$. This part is ok: You use non-emptiness of $H$ to produce an element $x\in H$, and then the criterion in the Proposition with $a=b=x$ to see that $e=xx^{-1}\in H$.  
Now before 3), I would say that you also have to show that for any $x\in H$, the inverse $x^{-1}$ is also in $H$? This follows by applying with $a=e$ and $b=x$: Then also $ex^{-1} = x^{-1}\in H$. If we have this, then in 3) we have that from $x$, $y\in H$ it follows $y^{-1}\in H$, and then from the criterion in the proposition with $a=x$ and $b=y^{-1}$, we see $xy=x(y^{-1})^{-1}=ab^{-1}\in H$.  
Finally, as "iff" means "if and only if", there is the other direction: when $H$ is a subgroup, then $ab^{-1}\in H$ for all $a$, $b\in H$... but this is easy.
Now for the other proof. It is true that a subgroup of an abelian group is itself abelian, but that fact is not needed here, and in the place where you use it, you still want to show that $H$ is a subgroup, so it doesn't make sense to apply it there!  
To show that $H$ is a subgroup, you want to use the proposition, so you have to show two things:


*

*$H$ is not empty. So you have to show that there is some $x\in G$ with $x^n = e$.  

*When $a$ and $b$ are in $H$, then also $ab^{-1}$ is in $H$. So assume that $a^n=e = b^n$, and you have to show that $(ab^{-1})^n=e$. (Here you will need that $G$ is abelian.) (Added later:) So it is not about showing $x^ny^{-n} = e$ or similar. Here your proof is unfortunately not correct. 
I can add more details for the last part, but I hope it is clearer now?  
