Suppose there is a group homomorphism $\phi: G \to G_0$. If $|G| = 1013$ and $|G_0 | = 55$, what can you say about the image of $G$ under $\phi$?
I get that cardinality of image($\phi$) should divide $55$, by Lagrange theorem. So it can be 1, 5, 11 or 55. How to proceed from here?
On
The order of $G$ is prime, so $G$ is simple. Therefore any group homomorphism from $G$ has kernel either $\{e\}$ or the whole $G$. In our case, the former is ruled out because $ϕ$ can't be injective (in fact $1013>55$, see pigeonhole principle); so, we are left with $\operatorname{ker}(ϕ)=G$ and every element of $G$ is mapped to the identity of $G_0$, namely $ϕ(G)=\{e_{G_0}\}$. (For this argument to work, the actual values of $|G|$ and $|G_0|$ are irrelevant, as long as $G$ is simple and $|G_0|<|G|$.)
For every element $g\in G$, let $|g|$ denote the order of $g$, that is the least positive integer $n$ such that $g^n=e$, where $e$ denotes the unit of $G$. Then we have $$\phi(g)^{|g|}=\phi(g^{|g|})=\phi(e)=e_0$$ where $e_0$ denotes the unit of $G_0$. We know also that, if $n$ is an integer such that $g^n=e$, then $|g|$ divides $n$. From our previuos remark we then deduce that $|\phi(g)|$ must divide $|g|$. Since $55$ and $1013$ are coprime, we conclude that $\phi(g)=e_0$.
EDIT: Let $n$ be an integer such that $g^n=e$. By the division Algorithm, we find integers $q, r$ such that $n=q|g|+r$, with $0\leq r<|g|$. Then we have $$g^r=g^{n-q|g|}=g^n\cdot(g^{|g|})^{-q}=e\cdot e=e$$ But since $|g|$ was least with that property, $r$ must be zero, hence $|g|$ divides $n$.