If G is a finite abelian group of order n and φ: G -> G’ is defined by φ(a)= a^m for all a element of G find the necessary and sufficient condition that φ be an isomorphism of G itself
Start by rewriting the question! You are asserting that "φ be an isomorphism of itself" which makes no sense. Grammatically you want "φ be an isomorphism of G to itelf", also called an "automorphism".
Also you have defined φ by "φ(a)= am" without saying what "m" is. A fixed member of G?
I think you meant that acts on an element by applying the group operation m times. Now, in order to say that this forms an automorphism, you have to show that
.
This is true because the group is Abelian. Next, for it to be an automorphism it has to be onto. Since the group is finite, if you can show that is one-to-one, it would imply that it is onto.
For it to be one-to-one, it is necessary that if , then it must be true that . Thus, implies that . That would definitely be true if , but that might be overkill. There might be a weaker condition that works.
It occurs to me that isn't enough. In the group , and at the same time , thus is not one-to-one even though . But the rest of my argument holds. I just can't think of a sufficient condition. Sorry
A condition that will absolutely work is if the order of the group is a prime number. then every element is a generator, which in turn means that the is a generator as long as . Can anyone think of a weaker condition?