Homomorphism $\phi: G \rightarrow H$, $\phi$ surjective, $\exists a \in H : |a| = 5$, show $\exists x \in G : |x| = 5$, where $|G|$ is finite.
I'm not sure if this proof is correct, but here's what I have so far:
By the First Isomorphism Theorem: $G/\ker\phi$ isomorphic to $\phi(G)$.
We know $\phi(e_G) = e_H$, so $e_G \in \ker\phi$.
So that means $\exists ge_G \in G/\ker\phi$ with a bijection mapping to the element of order $5$ in $H$, so that means $|ge_G| = 5 = |g|$ for some $g \in G$.
Notice, I did not use the fact that $\phi$ is surjective, meaning there's most likely something wrong with the proof.
It is not true that $ge_G\in G/\ker\phi$ because the elements of $G/\ker\phi$ are the cosets of the subgroup $\ker\phi$ in $G$. So for example $g+\ker\phi\in G/\ker\phi$.
Also note that you did use the fact that $\phi$ is surjective, when you take an element of $G/\ker\phi$ that maps to the element of order $5$ in $H$; such an element exists because $\phi$ is surjective.
But instead of looking at the quotient, consider the preimage of the element $a\in H$ with $|a|=5$. What can you say about its order?