Show a power set to be a subgroup of another power set 
Let $C$ and $D$ be sets with $C \subseteq D$. Prove the power set of $C$ - $P(C)$ - is a subgroup of $P(D)$.

I'll assume that the operation on $P(C)$ is $+$.
Also, I'll use the symmetric difference of two sets as the definition of $+$.
$A + B = (A - B) \cup (B - A)$.
$B = A^{-1}$ satisfies $(A - B) \cup (B - A) = \emptyset $, so $A^{-1} = A$ and $e = \emptyset$. If $A \in P(C)$, then $A^{-1} \in P(C)$. Every set contains $\emptyset$.
Suppose $A, B \in P(C)$. Then, $A + B = (A - B) \cup (B - A)$. Intuitively, a union of two sets in a set $S$ is in the set $S$. If this is true which I am not sure how to show we could(?) conclude that $P(C)$ is closed under addition.
How do I fix/improve this proof?
 A: You have the right idea - you give a perfectly valid proof that $P(C)$ is closed under inverses (and contains the identity), and you have the right idea to show it is closed under addition. All you need to show is that $A+B=(A-B)\cup (B-A)$ is a subset of $C$ given that $A$ and $B$ are, and that establishes that $P(C)$ is closed under addition - and you have the right idea that $A+B$ is the union of two subsets of $C$ and is hence a subset of $C$.
There's not a whole lot more than definitions to show that - I would think taking it on intuition that the union of two subsets of $C$ is a subset of $C$ would be fine. However, if you want proof, note the statement that $X\subseteq C$ and $Y\subseteq C$ can be written as

$$(x\in X \rightarrow x\in C)\wedge(x\in Y \rightarrow x\in C)$$

and the statement that $X\cup Y \subseteq C$ is equivalent to

$$(x\in X \vee x\in Y \rightarrow x\in C)$$

which is equivalent to the first statement, since we have two premises implying the same result, meaning either premise suffices to establish the result.
A: Your argument is right - all operations on the subsets of $C$ remain in $C$. Note that if $D$ is finite say $|D|=n$, then $(P(D), +) \cong (\mathbb{Z}/2\mathbb{Z})^n$, a direct sum of $n$ copies. This generalizes to infinite cardinals. 
