Let $G$ be abelian group and $\varphi:G\rightarrow G $ ,$\varphi(a)=a^{n},n\in \mathbb{N},n>1$
I need to find if $\varphi$ is homomorphism, and if so to find $ker(\varphi)$ and to tell if $\varphi$ is one-to-one and\or onto.
Attemept:
$\varphi(ab)=(ab)^n=(ab)(ab)\cdot \cdot\cdot n$ times $=a(ba)b\cdot \cdot \cdot $
$G$ is abelian so
$=(a\cdot a\cdot a\cdot \cdot \cdot n $ times$)( b\cdot b\cdot b\cdot \cdot \cdot n $ times$)$
$\varphi(a)\varphi(b)=a^{n}b^{n}=(a\cdot a\cdot a\cdot \cdot \cdot n$ times$)(b \cdot b\cdot \cdot \cdot n$ times $)$ $\Longrightarrow \varphi$ is homomorphism
Observe that $\varphi$ is a homomorphism, by your attempt, because the group is Abelian and hence $$ \varphi(ab)=(ab)^n=a^nb^n=\varphi(a)\varphi(b) $$ For the kernel observe that $a\in ker(\varphi)$ iff the order of $a$ devides $n$. On the other hand $\varphi$ is one to one iff $G$ has no element of order dividing $n$.
If $G$ is finite then $\varphi$ is one-to-one iff $(|G|,n)=1$, because if $p|(|G|,n)$, then there is an element of $G$ of order $p$, by Cauchy theorem, and hence $\varphi(a)=\varphi(e)$.