The binary operation $*$ for combining sets is defined as $A*B=(A\cup B)-(A\cap B)$.

To prove that $G$, consisting of the set of all subsets of a set $S$ (including the empty set and the set $S$ itself), together with the binary operation $*$, forms a group (assuming that the associative property is satisfied) it has to be shown that $G$ is closed, it contains an identity element and for each element in $G$ there is an inverse element contained in $G$.

If $A$ and $B$ are two subsets of the set $S$, then $A*B=(A\cup B)-(A\cap B)$ is also a set, and $(A\cup B)-(A\cap B)$ is the subset of $S$ shown in colour in Andrei's diagram. Hence the closure property is satisfied.

The union of any set $A$ and the empty set $\varphi$, is $A$, and there is no intersection between $A$ and $\varphi$ as $\varphi$ contains no elements to intersect.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{A*\varphi }& =(A\cup \varphi )-(A\cap \varphi )\\ \multicolumn{1}{c}{}& =A-\varphi \\ \multicolumn{1}{c}{}& =A\end{array}.}$

Therefore $\varphi$ is the identity element.

The intersection of a set with itself is itself, and the union of a set with itself is itself, for any set $A$, that is

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{A*A}& =(A\cup A)-(A\cap A)\\ \multicolumn{1}{c}{}& =A-A\\ \multicolumn{1}{c}{}& =\varphi \end{array}.}$

Therefore the inverse of any element is itself, each element of $G$ is self inverse.

The fourth property, associativity, was assumed so we have shown $G$ is a group.

To solve $\{1,2,4\}*X=\{3,4\}$, rewrite it as $A*X=B$, where the solution is $X=A^{-1}*B$. We consider the set of all subsets of the natural numbers and note that this is an example of the group discussed above. Hence as the element $\{1,2,4\}$ is self inverse:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{X}& =\{1,2,4\}*\{3,4\}\\ \multicolumn{1}{c}{}& =\{1,2,4\}\cup \{3,4\}-\{1,2,4\}\cap \{3,4\}\\ \multicolumn{1}{c}{}& =\{1,2,3,4\}-\{4\}\\ \multicolumn{1}{c}{}& =\{1,2,3\}\end{array}.}$

Check:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\{1,2,4\}*\{1,2,3\}}& =\{1,2,4\}\cup \{1,2,3\}-\{1,2,4\}\cap \{1,2,3\}\\ \multicolumn{1}{c}{}& =\{1,2,3,4\}-\{1,2\}\\ \multicolumn{1}{c}{}& =\{3,4\}.\end{array}}$