Given any abelian group $A$, let $T(A)$ be its torsion subgroup and $F(A)=A/T(A)$. A homomorphism $\varphi:A\to B$ induces $T(\varphi):T(A)\to T(B)$, $F(\varphi):F(A)\to F(B)$, and $\varphi$ is one-to-one iff $T(\varphi)$ and $F(\varphi)$ are one-to-one. And the injection $\varphi$ is essential, iff $T(\varphi)$ and $F(\varphi)$ are essential. Now suppose given
where $A\subset C$ is to be regarded as fixed, and $B$ is an essential extension of $A$. It's readily to see that if $T(B)$ and $F(B)$ are maximal, so is $B$. However, the converse is not true: If $B$ is maximal, so is $T(B)$, but $F(B)$ may fail to be maximal.
I read this from GTM4 and I didn't know how to find a counter example that $F(B)$ is not maximal.
Here's one counter example:
$C=\mathbb{Z}\oplus\mathbb{Z}_2$
$A$ is the subgroup of $C$ generated by $(2,1)$ where $2 \in \mathbb{Z}, 1 \in \mathbb{Z}_2$.
Then we can observe that $A$ has no nontrivial essential extensions in $C$: Clearly $(0, 1)$ cannot be in any essential extension of $A$. For any nontrivial essential extension, we must have $F(B) \neq F(A) $ considered as submodules of $F(C)$. Hence $F(B)=F(C)$, there exists $(1, a) \in B$. Then $(2, 0) \in B$, and $(0, 1) \in B$, which is impossible.
But $F(A)=2F(C)$ has $F(C)$ as its maximal essential extension.
This example is constructed by playing with subgroups of a f.g. abelian group that is not obtained by a direct sum of subgroups of the torsion component and the free component.