Copyright © University of Cambridge. All rights reserved.
Thanks Jack from Pate's School, Yosef from Yeshivat Rambam High
School, Baltimore, Andrei from Tudor Vianu National College,
Bucharest, Romania and Curt from Reigate College for your
solutions.
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 $\phi$, is $A$, and
there is no intersection between $A$ and $\phi$ as $\phi$ contains
no elements to intersect.
$$\eqalign{ A*\phi &= (A\cup \phi)-(A\cap \phi)\cr &= A -
\phi \cr &= A}.$$
Therefore $\phi$ 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
$$\eqalign{ A*A &= (A\cup A) - (A\cap A) \cr &= A - A \cr
&= \phi }.$$
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:
$$\eqalign{ X &= \{1,2,4\}*\{3,4\} \cr &=
\{1,2,4\}\cup\{3,4\}- \{1,2,4\}\cap \{3,4\} \cr &= \{1,2,3,4\}
- \{4\} \cr &= \{1,2,3\} }.$$ Check: $$\eqalign{
\{1,2,4\}*\{1,2,3\}&= \{1,2,4\}\cup\{1,2,3\} -
\{1,2,4\}\cap\{1,2,3\}\cr &=\{1,2,3,4\}- \{1,2\} \cr &=
\{3,4\} .}$$