How to compute identity from symmetric difference?

Question

We know that $(P(X), \Delta) $ forms a group, where $P(X) $ is the power set on the non empty set $X$ and $A\Delta B=(A-B) \cup (B-A)$ for all $A, B\in P(X) $. Clearly $\emptyset$ is the identity element. How can i compute $E$ from the relation $Y\Delta E=Y$ for all $Y\in P(X) $.

Answer 1

$$ E= E\Delta E= (E\setminus E) \cup (E\setminus E)= \emptyset\cup\emptyset= \emptyset. $$

Answer 2

By definition, we have $B-A = \{x|x \in B \text{ and }x \notin A\}$. Also notice that since $A\Delta B=(A-B) \cup (B-A)$, we have $B-A \subseteq A \Delta B$.

Now, we are looking for the set $E$ that satisfies $Y\Delta E= (Y-E) \cup (E-Y) = Y$ for all $Y\in P(X)$. First of all, if $E-Y \ne \emptyset$, then there exists an element $a$ such that $a \in E$ but $a \notin Y$. But then, since $E-Y \subseteq Y$, we have a contradiction because $a \in E-Y$ but $a \notin Y$. So we need to have $E-Y = \emptyset$. This means that $E \subseteq Y$.

Now, since $E-Y = \emptyset$, we have $Y \Delta E = (Y-E) \cup (E-Y) = Y-E = Y$ for all $Y \in P(X)$. And this is possible only when $E = \emptyset$.