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 $\displaystyle \phi$ 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
$\displaystyle (ab)^m = \phi(ab) = \phi(a)\phi(b) = a^m b^m$.
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 $\displaystyle \phi$ is one-to-one, it would imply that it is onto.
For it to be one-to-one, it is necessary that $\displaystyle \forall a,b \in G$ if $\displaystyle \phi(a) = \phi(b)$, then it must be true that $\displaystyle a=b$. Thus, $\displaystyle a^m = b^m$ implies that $\displaystyle a=b$. That would definitely be true if $\displaystyle (m,n)=1$, but that might be overkill. There might be a weaker condition that works.
It occurs to me that $\displaystyle (m,n)=1$ isn't enough. In the group $\displaystyle U(9)$, $\displaystyle 4^2 = 16 = 7 (mod 9)$ and at the same time $\displaystyle 5^2 = 25 = 7 (mod 9)$, thus $\displaystyle \phi(a) = a^2$ is not one-to-one even though $\displaystyle (2,9) = 1$. 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 $\displaystyle \phi(a) = a^m$ is a generator as long as $\displaystyle m < n$. Can anyone think of a weaker condition?