Critique of this subgroup proof? This is an exercise out of a Dover book on abstract algebra. I was hoping I could get some feedback on this proof, as I'm still fairly new to this level of proof writing. Thanks! Here's the exercise:
Let $G$ be an Abelian group whose operation is multiplication, let $H$ be a subgroup of $G$, and let $K = \{x \in G:$ for some positive integer $ n$, $x^n \in H \}$. Prove that $K$ is a subgroup of $G$.
My proof:
To show that $K$ is a subgroup of $G$, it is sufficient to show that $K$ is a subset of $G$ and that $K$ is closed under multiplication and inverses. The first condition is trivially satisfied, because $K$ consists only of elements of $G$. We now show that $K$ is closed under multiplication.
Let $x,y \in K$. We know $x,y \in G$ and $x^n,y^m \in H$ for some positive integers $m,n$. Additionally, $x^m,y^n \in H$. Since $x,y \in G$ and $G$ is closed under multiplication, $x \cdot y \in G$. Similarly, $x^n \cdot x^m \cdot y^n \cdot y^m \in H$. Now $(x \cdot y)^{n+m} = \underbrace{(x \cdot y) \cdot (x \cdot y) \cdots (x\cdot y)}_\text{n+m}$. Since $G$ is Abelian, we can rearrange this as $\underbrace{x \cdot x \cdots x}_\text{n+m} \cdot \underbrace{y \cdot y \cdots y}_\text{n+m} \\ = \underbrace{x \cdot x \cdots x}_\text{n} \cdot \underbrace{x \cdot x \cdots x}_\text{m} \cdot \underbrace{y \cdot y \cdots y}_\text{n} \cdot \underbrace{y \cdot y \cdots y}_\text{m} = x^n \cdot x^m \cdot y^n \cdot y^m \in H$. Therefore, since $x \cdot y \in G$, $n+m$ is a positive integer, and $(x \cdot y)^{n+m} \in H$, $x \cdot y \in K$. Therefore, $K$ is closed with respect to multiplication.
Now we prove that $K$ is closed with respect to inverses. Let $x \in K$. Therefore $x \in G$ and $x^n \in H$ for some positive integer $n$. Since $x \in G$ and $G$ is closed to inverses, $\frac{1}{x} \in G$. Similarly, $\frac{1}{x^n} \in H$. $(\frac{1}{x})^n = \frac{1}{x^n} \in H$. Therefore, $\frac{1}{x} \in K$, and $K$ is closed with respect to inverses. Thus, $K$ is a subgroup of $G$.
 A: You write: 'Let $x,y∈K$. We know $x,y∈G$ and $x^n,y^m∈H$ for some positive integers $m,n$. Additionally, $x^m,y^n∈H$'.
This is not always true! Consider the group $8\mathbb{Z} \subset \mathbb{Z}$. Note that if $x = 2$ and $y = 4$, then $4 + 4 \in 8 \mathbb{Z}$ and $2 + 2 + 2 +2 \in 8 \mathbb{Z}$, but $2 +2 \not \in 8 \mathbb{Z}$. 
Instead of looking at $(x\cdot y)^{m+n}$, try looking at $(x \cdot y)^{m \cdot n}$. 
Hope my hint help! Best of luck!
A: Just one nitpick. 
Third sentence, third paragraph: The fact that $x^m$ and $y^n$ are in $H$ only gives you the fact that $(x^m \cdot y^n) \in H$. There is no reason why $x^n \cdot x^m \cdot y^n \cdot y^m$ should be in $H$. But if $x^m$ and $y^n$ are in $H$ then any power of these elements are also in $H$. So try $(x \cdot y)^{mn}$. 
The fact that the inverse of any element in $H$ is also in $H$ seems to be proven correctly. 
A: Your proof for the inverses is fine.  
Are you sure about the statement that x^{m} and y^{n} are also in H?  Might want to double-check how you know that.
Also, you do not need to note that xy is in G; it is not relevant to proving the closure of K under multiplication.  Your argument using m+n rests on the statement that I said you should check on.
Try nm instead of n+m.
A: The other comments have pointed out a mistake, but I think everything else looks good :)
One thing though, although it's obvious, it's important to state why the identity element is in $K$. Without the identity element, $K$ isn't a subgroup.
A: There is no reason $x^m$ or $y^n$ should be in $H$. So you need to change the proof that $xy\in K$.
Also you may prove directly that if $x,y\in K$ then $xy^{-1}\in K$ which can prove the same subgroup property in one step.
