First to get clear about the set $G$: it is the set of all subsets of a set $A$. (So the elements of $G$ are sets.) And by definition, $G$ is therefore the powerset of $A$, denoted $G =\mathcal{P}(A))$. So $|G| = 2^{|A|}$, which is finite if and only if $|A|$ is finite, and is infinite otherwise.
The operation on the sets $g_1, g_2 \in G$ is the symmetric difference of $\;g_1\;$ and $\;g_2\;$ which we'll denote as $\;g_1\;\triangle\;g_2\;$ and is defined as the set of elements which are in either of the sets but not in their intersection: $\;g_1\;\triangle\;g_2 \;= \;(g_1\cup g_2) - (g_1 \cap g_2)\tag{1}$
Hence, the symmetric difference $g_i \;\triangle\; g_j \in G\;$ for all $\;g_i, g_j \in G$. ($G$ is the set of ALL subsets of $A$, so it must include any possible set resulting from symmetric difference between any arbitrary sets in $G$, which are also subsets of $A$). That is we have now established, that the symmetric difference is closed on $G$.
The power set $G$ of any set $A$ becomes an abelian group under the operation of symmetric difference:
Why abelian? Easy to justify, just use the definition in $(1)$ above: it's defined in a way that $g_1 \triangle g_2$ means exactly the same set as $g_2 \triangle g_1$, for any two $g \in G$.
As you note, the symmetric difference on $G$ is associative, which can be shown using the definition in $(1)$, by showing for any $f, g, h \in G, (f\; \triangle\; g) \triangle \;h = f\;\triangle\; (g \;\triangle\; h)$.
The empty set is the identity of the group (it would be good to justify this this, too), and
every element in this group is its own inverse. (Can you justify this, as well? Just show for any $g_i \in G, g_i\;\triangle \; g_i = \varnothing$).
The justifications for these properties is very straightforward, but good to include for a proof that the symmetric difference, together with the set $G$ as defined, form an abelian group.
So, you've covered most of the bases, but you simply want to confirm/note the closure of the symmetric difference operation on $G$ and why, add a bit of justification for the identity and inverse claims, and to address whether, or when, $G$ is finite/infinite: this last point being that the order of $G$ depends on the cardinality of $A$.
Since you already solve the problem, this answer should play as a comment but I put it here for proper format.
$P_D$ can be identified with the set of functions
$$f:D\to \Bbb Z/2,$$
from $D$ to the integers mod $2$, where each subset $A$ is identified to its characteristic function $1_A$ such that $$1_A(a)=1\iff a\in A.$$
Then the operation $A+B$ translates to $1_A+1_B$ in normal sense.
It follows easily that
- the identity is the zero function, i.e. the characteristic function of the empty set, and
- the inverse of any function is itself.
Best Answer
You need to show that for any $X, Y \in \mathcal P(H),\, X\Delta Y\in\mathcal P(H)$.
If we get rid of the powerset notation, the above is equivalent to: $$\forall X, Y \subseteq H: X\Delta Y\subseteq H$$