Question

Show that the mapping \(\phi: S_{3} \rightarrow S_{3}\) defined by letting \(\phi(x)=x^{-1}\) for all \(x \in S_{3}\) is not an automorphism of \(S_{3}\).

Short answer

The map \( \phi(x) = x^{-1} \) is not an automorphism of \( S_3 \) since it fails to preserve the group operation.

Step-by-step solution

Define Automorphism

To start, let's define what it means for a mapping to be an automorphism. A mapping \( \phi : G \rightarrow G \) is an automorphism if it is a bijective homomorphism, meaning it must be both bijective (one-to-one and onto) and it must preserve the group operation. In the case of a symmetric group like \( S_3 \), \( \phi(xy) = \phi(x) \phi(y) \) must hold for all \( x, y \in S_3 \).

Verify Bijectivity

Check if \( \phi(x) = x^{-1} \) is bijective. In general, the inversion map is bijective: every element has a unique inverse, and each inverse is uniquely assigned to an element. Thus, \( \phi: S_3 \rightarrow S_3 \) is bijective.

Check for Homomorphism

Next, check if \( \phi \) is a homomorphism by verifying \( \phi(ab) = \phi(a) \phi(b) \) for all \( a, b \in S_3 \). Consider two permutations \( \sigma = (12) \) and \( \tau = (123) \). Calculate \( \phi(\sigma) = \sigma^{-1} = (12) \) and \( \phi(\tau) = \tau^{-1} = (132) \). Now calculate \( \phi(\sigma \tau) = \phi((13)) = (13) \) and \( \phi(\sigma) \phi(\tau) = (12)(132) = (23) \). Since \( (13) eq (23) \), \( \phi \) does not preserve the operation.

Conclusion on Automorphism

Since \( \phi \) does not preserve the group operation, it is not a homomorphism. Therefore, even though \( \phi \) is bijective, it is not an automorphism of \( S_3 \), because it fails to satisfy the homomorphism property.

Key concepts

Automorphism

An automorphism is a special kind of mapping within group theory. It is a function from a group to itself that preserves the group structure. For a function \( \phi: G \rightarrow G\) to qualify as an automorphism, it must satisfy two main criteria:
- It must be *bijective*. This means every element of the group is mapped to one unique element, and every element in the target group is the image of exactly one element from the domain.
- It must be a *homomorphism*. This means it should preserve the group operation; if you take two elements \( x \) and \( y \) from the group and operate on them, and then apply the automorphism function \( \phi \), the result should be the same as applying the function to \( x \) and \( y \) separately and then combining the results from \( \phi(x) \) and \( \phi(y) \).
In simple terms, while bijectivity ensures every element aligns uniquely, homomorphism ensures the entire "structure" of processing is preserved by the mapping. Without both conditions, the mapping \(\phi\) cannot be considered an automorphism.

Symmetric Group

The symmetric group, denoted as \( S_n \), is the group of all permutations of \( n \) objects. It is a key object of study in group theory. Such groups consist of all possible ways to order a set of \( n \) distinct objects, and the group operation is the *composition of permutations*.
For instance, \( S_3 \) involves 3 objects which can be ordered in \( 3! = 6 \) different ways. These permutations might look like \((1)(2)(3)\), \((12)\), \((13)\), and so forth. Each of these permutations is an element of the symmetric group \( S_3 \), and together they form the structure of this group.
Symmetric groups are crucial as they provide insight into the arrangement and symmetry of objects and many properties and results in group theory are first explored within these groups.

Bijective Homomorphism

A bijective homomorphism is a crucial concept in group theory. It is a mapping which holds two properties:
- **Bijective**: This property ensures that each element from the domain maps to a unique element in the codomain and that every element in the codomain is covered.
- **Homomorphic**: This property demands that the group operation is preserved under the mapping.
Put more simply, a bijective homomorphism not only forms a perfect one-to-one correspondence but also maintains the group operation, meaning if \( a, b \) belong to a group, then the operation \( (ab) \) under the mapping \(\phi\) will be equivalent to \(\phi(a)\phi(b)\).
This is critical when considering automorphisms, as any mapping that is a bijective homomorphism could potentially be an automorphism.

Inversion Map

In group theory, the inversion map is a fascinating concept. It is an operation that involves mapping elements of a group to their respective inverses. Essentially, for any element \( x \) in a group, the inversion map will assign the element \( x^{-1} \).
For example, if \( x = (12) \) is a permutation in the symmetric group \( S_3 \), its inverse under the inversion map would also be \( (12) \). This is because for permutations, the inverse is the same permutation that, when composed with the original, returns the identity element of the group.
In theory, the inversion map is always bijective because every element has a unique inverse, but it does not necessarily preserve the operation of non-Abelian groups like \( S_3 \). That is the reason why it might fail the homomorphism condition, and hence is often not an automorphism even though it is bijective. Understanding the intricacies of these operations helps clarify the structure of groups and their possible mappings.