Given the question:
If $G$ is a finite abelian group of order $n$ and $\varphi: G\to G$ is defined by $\varphi(a) = a^m$ for all $a \in G$, find the necessary and sufficient condition that $\varphi$ be an isomorphism of $G$ onto itself.
I could came up with the following "answer" but I am not sure, so wanted ask for some guidance.
It is easy to see that if $\gcd(m,n)=1$, then Kernel is only identity as $x^m=x^n=e$, for all elements in the group. Then as $\gcd$ is a linear combination, $x=e$ is attained.
For the hard side , I am trying to argue as follows. Say $x \in\ker\varphi$, then $\langle x\rangle $ is a subgroup of $\ker\varphi$, and say $p$ is a prime divisor of order of $\langle x\rangle$, then $\langle x^{{\rm order}(x)/p}\rangle $ has order $p$, then $\gcd (p,m)=1$, I think to make only $1$ element in kernel, as otherwise there are at least $p$ non identity elements in kernel. Now, this is as long as $p$ is a divisor of orders of elements in kernel. But is there any prime divisor, $q,$ of $n$ such that $q$ does not divide any order?
In this case, consider any element $x\neq e$, then $x^q\neq e$, as otherwise ${\rm order}(x)$ divides $q$ and as $q$ does not divide ${\rm order}(x)$, order is $1$, but $x$ was non identity. Now what this says is only solution of $x^{q}=e$ is $x=e$. Now $\forall x \in G$, $x^{{\rm order}(G)}=x^{\frac{{\rm order}(G)}q*q}=e$ , hence $x^{\frac{{\rm order}(G)}q}=e$ for all $x$. Now, this seems to me like a very strong claim but i can not make use of it, if this is useful.
The "hard side" can be done more directly. Let's suppose that $\gcd(m,n) \neq 1$. Then there exists a prime number $p$ such that $p|n$ and $p|m$. By Cauchy's theorem, there exists $a \in G$ that has order $p$. Hence you have $a \neq e$ and $a^m = e$, so $\varphi$ is not injective.