The proposition is as follows:" Let $G$ be a finite Abelian group. Prove that $\varphi_k(g) = g^k$ is an endomorphism of $G$. If $\varphi_k$ is an automorphism of $G$, then $k$ and $|G|$ are coprime. "
I have already proven the endomorphism part, but I'm facing difficulties with the latter part of the proposition. I've attempted to use a proof by contradiction, trying to construct an element $g^{\frac{m}{(m,k)}}$, where $m$ is the order of the element $g$, in order to show that this mapping is not injective and leads to a contradiction. However, at the moment, $(m, k)$ could be $1$, and I'm unsure how to proceed.
I have seen the answer in Conditions for $x \mapsto x^k$ to be an automorphism but at this time I haven't studied neither Cauchy's theorem nor cyclic groups.
I did a proof without using Cauchy's theorem or theory of cyclic subgroups, as the OP wanted.
If $|G|=1$ then the result is trivial. We assume then that $|G|\not=1$. Suppose that there exists a prime divisor $p$ of $k$ which also divides $|G|$. Now, since $\varphi_k$ is an automorphism it follows that $\varphi_p$ is injective. To prove it let $g,h\in G$ such that $g^p=h^p$. Then $g^k=(g^p)^{k/p}=(h^p)^{k/p}=h^k$ and hence $g=h$. Thus, each $\varphi_p$ is injective. Since the composition of injections is an injection we have that $\varphi_{p^m}$ is also injective for each natural $m$.
We consider $p^m$ to be the maximum power of $p$ dividing $|G|$. If we take $g\in G$ then $$1^{p^m}=1=g^{|G|}=g^{\frac{|G|}{p^m}p^m}=(g^{\frac{|G|}{p^m}})^{p^m}.$$ Since $\varphi_{p^m}$ is injective we deduce that $g^{\frac{|G|}{p^m}}=1$. Denoting the order of an element $h\in G$ by $o(h)$, then it follows that $o(g)\mid\frac{|G|}{p^m}$ and hence $p\not\mid o(g)$ by the choice of $m$. Then, we shall prove that $$p\not\mid o(g)\text{ for each }g\in G\implies p\not\mid|G|,$$ or equivalently, $$p\mid|G|\implies\exists g\in G:p\mid o(g).$$ This will contradict the assumption that $k$ and $|G|$ have a common prime divisor $p$. We consider the set $T=\{g\in G: p\not\mid o(g)\}$. If $T=\{1_G\}$ we are done. In other case, we take $1_G\not=g\in T$ of minimal order. If $n\in\mathbb Z$, then $(g^n)^{o(g)}=(g^{o(g)})^n=1$ and therefore $o(g^n)\mid o(g)$. We affirm that there exists $n\in\mathbb Z$ such that $p\mid o(g^n)$. If this does not occur then for each $n\in\mathbb Z$ it would hold $p\not\mid o(g^n)\leq o(g)$. The minimality of the order of $g$ implies in this case that $o(g^n)=o(g)$ for every $n\in\mathbb Z$. Taking $n=o(g)$ we obtain a contradiction, which concludes the proof.