91Ó°ÊÓ

In group theory, closure is a fundamental concept. For a set to be a group, the group operation must be closed within the set. This means that performing the operation on any two elements of the set should produce another element in the same set.

In the given problem, we have transformations of the form \( \vartheta_{a,b}(x) = ax + b \). To verify closure, consider two such maps, say \( \vartheta_{a,b} \) and \( \vartheta_{c,d} \). Their composition is \( (\vartheta_{a, b} \circ \vartheta_{c, d})(x) = a(cx + d) + b = acx + ad + b \), which can be rewritten as another map of the form \( \vartheta_{ac, ad+b}(x) \).

Since \( ac eq 0 \) (because \( a eq 0 \) and \( c eq 0 \)), this new map is also in \( G \). Therefore, \( G \) is closed under composition, satisfying the closure property of a group.

The identity element is a special element in a group that, when used in the group operation with any other element, returns the other element unchanged.

For the set \( G \) of transformations \( \vartheta_{a,b}(x) = ax + b \), the map \( \vartheta_{1,0}(x) = x \) is the identity element. This is because, when you compose any other map \( \vartheta_{a,b} \) with \( \vartheta_{1,0} \), the result is \( \vartheta_{1,0} \circ \vartheta_{a,b}(x) = ax + b \) and \( \vartheta_{a,b} \circ \vartheta_{1,0}(x) = ax + b \).

Thus, \( \vartheta_{1,0} \) does not alter any map \( \vartheta_{a,b} \), confirming its role as the identity element in \( G \).

In a group, each element must have an inverse, meaning there should be another element in the group that reverses the effect of the original element through the group operation.

For an element \( \vartheta_{a,b} \) in \( G \), the inverse map \( \vartheta_{a, b}^{-1}(x) \) must satisfy the condition that \( (\vartheta_{a, b} \circ \vartheta_{a, b}^{-1})(x) = x \). From the equation \( ax + b = y \), solving for \( x \) gives \( x = \frac{y - b}{a} \). Therefore, the inverse function is \( \vartheta_{a^{-1}, -\frac{b}{a}}(x) = \frac{x - b}{a} \).

This map is also of the required form, ensuring each transformation in \( G \) has its inverse, reaffirming the group property of having inverses for all elements.

Associativity is a key characteristic of any group, requiring that the composition of operations is independent of how the operations are grouped. This means \( (f \circ g) \circ h = f \circ (g \circ h) \) for any functions \( f, g, \) and \( h \) in the group.

For our transformation group \( G \), consider three elements \( \vartheta_{a,b}, \vartheta_{c,d}, \) and \( \vartheta_{e,f} \). Their compositions yield the same result whether we compute \( \vartheta_{a,b} \circ (\vartheta_{c,d} \circ \vartheta_{e,f}) \) or \( (\vartheta_{a,b} \circ \vartheta_{c,d}) \circ \vartheta_{e,f} \), both simplifying to \( a(cex + (cd+f)) + b \).

This proves the associativity condition, thus supporting \( G \)'s status as a group with respect to function composition.

In an Abelian group, the order in which you perform the group operation does not matter; that is, it is commutative. A subgroup of a given group is Abelian if it satisfies this property.

Within \( G \), consider the subset \( \{\vartheta_{1,e}: e \in \mathbb{R}\} \), where each transformation is of the form \( x + e \). These transformations commute under composition, as shown by \( (\vartheta_{1, e} \circ \vartheta_{1, f})(x) = x + e + f = (\vartheta_{1, f} \circ \vartheta_{1, e})(x) \).

This subset not only possesses an identity element, \( \vartheta_{1,0} \), but also for each element \( \vartheta_{1,e} \), there exists an inverse \( \vartheta_{1,-e} \). Therefore, this subset is an Abelian subgroup of \( G \), where all operations commute and every element and its inverse are accounted for.