Problem 2
Question
Prove that each of the following sets, with the indicated operation, is an abelian group. $$ x * y=\frac{x y}{2}, \text { on the set }\\{x \in \mathbb{R}: x \neq 0\\} $$
Step-by-Step Solution
Verified Answer
The set with the operation is an abelian group because it satisfies closure, associativity, having an identity and inverses, and commutativity.
1Step 1: Check Closure
We need to show that for any two elements \(x, y\) in the set, the result of the operation \(x * y = \frac{xy}{2}\) is also in the set.If \(x eq 0\) and \(y eq 0\), then \(xy eq 0\). Since \(xy eq 0\), \(\frac{xy}{2} eq 0\). Therefore, \(\frac{xy}{2}\) is an element of the set \(\{z \in \mathbb{R} : z eq 0\}\). Thus, the set is closed under the operation.
2Step 2: Check Associativity
We must show that the operation * is associative, meaning that for any \(x, y, z\) in the set, \((x * y) * z = x * (y * z)\).Calculate:\((x * y) * z = \left(\frac{xy}{2}\right) * z = \frac{(\frac{xy}{2})z}{2} = \frac{xyz}{4}\).Calculate:\(x * (y * z) = x * \left(\frac{yz}{2}\right) = \frac{x(\frac{yz}{2})}{2} = \frac{xyz}{4}\).Since both expressions simplify to \(\frac{xyz}{4}\), the operation * is associative.
3Step 3: Check Existence of Identity Element
Find an element \(e\) in the set such that for every element \(x\) in the set, \(x * e = e * x = x\).Solve \(x * e = \frac{xe}{2} = x\) for \(e\):\(\frac{xe}{2} = x\) implies \(e = 2\).Check that when \(e = 2\), \(x * e = \frac{xe}{2} = x\) and \(e * x = \frac{ex}{2} = x\).Thus, the identity element is \(e = 2\).
4Step 4: Check Existence of Inverses
For each element \(x\) in the set, find an element \(x^{-1}\) such that \(x * x^{-1} = x^{-1} * x = e\), where \(e\) is the identity element found in Step 3.Solve \(x * x^{-1} = \frac{xx^{-1}}{2} = 2\):\(\frac{xx^{-1}}{2} = 2\) implies \(xx^{-1} = 4\), so \(x^{-1} = \frac{4}{x}\).Thus, every element \(x\) has an inverse \(x^{-1} = \frac{4}{x}\) in the set.
5Step 5: Check Commutativity
Finally, show that the operation is commutative, i.e., for any \(x, y\) in the set, \(x * y = y * x\).Calculate both expressions:\(x * y = \frac{xy}{2}\) and \(y * x = \frac{yx}{2}\).Since \(xy = yx\), \(x * y = y * x\), and the operation is commutative.
Key Concepts
Group theoryAssociativityIdentity elementInverse element
Group theory
Group theory is a branch of mathematics that studies algebraic structures known as groups. In group theory, a group consists of a set of elements along with an operation that combines any two elements to form a third element.
The group must satisfy four essential properties for the set and operation:
The group must satisfy four essential properties for the set and operation:
- Closure: Combining any two elements in the set results in another element from the same set.
- Associativity: The operation must be associative.
- Identity element: There must exist an element in the set that, when used in the operation with any element of the set, leaves the element unchanged.
- Inverse element: Every element in the set must have an inverse, such that combining the element with its inverse using the operation results in the identity element.
Associativity
Associativity is a crucial property in group theory. It requires that, while performing an operation on any three elements in a set, the order of performing the operations does not affect the final result. In other words, changing the grouping of the operations does not alter the outcome.
Mathematically, for any elements \(a, b, c\) in a group, the expression \((a * b) * c\) must equal \(a * (b * c)\).
Let's look at how associativity was verified for the original exercise's operation \(*\), defined by \(x * y = \frac{xy}{2}\):
Mathematically, for any elements \(a, b, c\) in a group, the expression \((a * b) * c\) must equal \(a * (b * c)\).
Let's look at how associativity was verified for the original exercise's operation \(*\), defined by \(x * y = \frac{xy}{2}\):
- Considering three elements \(x, y, z\), we calculate \((x * y) * z\), resulting in \(\frac{xyz}{4}\).
- Then, we calculate \(x * (y * z)\), also resulting in \(\frac{xyz}{4}\).
Identity element
The identity element is a special element in a group that, when used in the operation with any element of the group, does not change that element. It's like a "do nothing" element in the world of groups.
In our specific exercise, we want to find an identity element \(e\) such that when combined with any element \(x\) using \( * \), it returns \(x\) unchanged: \(x * e = e * x = x\).
Given the operation \(x * y = \frac{xy}{2}\), we solved for \(e\) using \(x * e = \frac{xe}{2} = x\).
This equation simplifies to give us \(e = 2\).
Ensuring that when \(e = 2\), \(x * e\) and \(e * x\) both return \(x\), confirms that \(2\) serves as the identity element in our group.
In our specific exercise, we want to find an identity element \(e\) such that when combined with any element \(x\) using \( * \), it returns \(x\) unchanged: \(x * e = e * x = x\).
Given the operation \(x * y = \frac{xy}{2}\), we solved for \(e\) using \(x * e = \frac{xe}{2} = x\).
This equation simplifies to give us \(e = 2\).
Ensuring that when \(e = 2\), \(x * e\) and \(e * x\) both return \(x\), confirms that \(2\) serves as the identity element in our group.
Inverse element
An inverse element in group theory is one that can "reverse" the operation with another element to yield the identity element. For every element \(x\) in a group, there should be an inverse \(x^{-1}\) such that their combination produces the identity element \(e\).
In the original exercise, this means finding \(x^{-1}\) such that \(x * x^{-1} = x^{-1} * x = 2\), where \(2\) is the identity element.
Substituting \(x * x^{-1} = \frac{xx^{-1}}{2} = 2\), we determine that \(xx^{-1} = 4\).
Solving this equation gives us the inverse element \(x^{-1} = \frac{4}{x}\).
This step ensures that every element has an inverse in our set, satisfying one of the core components of a group structure.
In the original exercise, this means finding \(x^{-1}\) such that \(x * x^{-1} = x^{-1} * x = 2\), where \(2\) is the identity element.
Substituting \(x * x^{-1} = \frac{xx^{-1}}{2} = 2\), we determine that \(xx^{-1} = 4\).
Solving this equation gives us the inverse element \(x^{-1} = \frac{4}{x}\).
This step ensures that every element has an inverse in our set, satisfying one of the core components of a group structure.
Other exercises in this chapter
Problem 1
Show that \(\left(a_{1}, a_{2}, \ldots, a_{n}\right)+\left(b_{1}, b_{2}, \ldots, b_{n}\right)=\left(b_{1}, b_{2}, \ldots, b_{n}\right)+\left(a_{1}, a_{2}, \ldot
View solution Problem 1
The symbol \(\mathbb{R} \times \mathbb{R}\) represents the set of all ordered pairs \((x, y)\) of real numbers. \(\mathbb{R} \times \mathbb{R}\) may therefore b
View solution Problem 3
Show that \(\left(a_{1}, \ldots, a_{n}\right)+\left[\left(b_{1}, \ldots, b_{n}\right)+\left(c_{1}, \ldots, c_{n}\right)\right]=\left[\left(a_{1}, \ldots, a_{n}\
View solution Problem 3
If \(A\) and \(B\) are any two sets, their symmetric difference is the set \(A+B\) defined as follows: $$ A+B=(A-B) \cup(B-A) $$ The shaded area is \(A+B\) It i
View solution