If S contains at least 2 elements(denoted x and y), then among the set of all mappings from S into S, there exists a mapping which to x associates x and to y associates x as well.
Therefore, this mapping is not injective.
So its inverse doesn't exist, which contradicts the fact that the set to which it belongs should be a group.
Hence, S cannot but contain only one element. And under this condition we can prove easily that the set of all mappings from S into S contains also only one element: the identity fuction, so it's a group.
To conclude, the set of all mappings from S into S forms a group iff S contains only one element.
2.
This property is included in the definition of a field.
Answers & Comments
Verified answer
1.
If S contains at least 2 elements(denoted x and y), then among the set of all mappings from S into S, there exists a mapping which to x associates x and to y associates x as well.
Therefore, this mapping is not injective.
So its inverse doesn't exist, which contradicts the fact that the set to which it belongs should be a group.
Hence, S cannot but contain only one element. And under this condition we can prove easily that the set of all mappings from S into S contains also only one element: the identity fuction, so it's a group.
To conclude, the set of all mappings from S into S forms a group iff S contains only one element.
2.
This property is included in the definition of a field.