I want to prove the following:
Let $G$ be a finite abelian $p$-group that is not cyclic. Let $L \ne {1}$ be a subgroup of $G$ and $U$ be a maximal subgroup of L then there exists a maximal subgroup $M$ of $G$ such that $U \leq M$ and $L \nleq M$.
Proof. If $L=G$ then we are done. Suppose $L \ne G$ . Let $|G|=p^{n}$ then $|L|=p^{n-i}$ and $|U|=p^{n-i-1}$ for some $0<i<n$. There is $x_{1} \in G$ such that $x_{1} \notin L$. Thus $|U\langle x_{1}\rangle|=p^{n-i}$ and does not contain L. There is $x_{2} \in G$ such that $x_{2} \notin L$ and $x_{2} \notin |U\langle x_{1}\rangle|$. Thus $|U\langle x_{1}\rangle\langle x_{2}\rangle|=p^{n-i+1}$. Continuing like this, we get $|U\langle x_{1}\rangle \langle x_{2}\rangle\cdots \langle x_{i}\rangle|=p^{n-1}$ is a maximal subgroup of $G$. The problem is, I am not sure that this subgroup does not contain $L $.
Thanks in advance.
Your proof fails if G is cyclic, by the uniqueness of sbgps. of a given divisor of the order of the group.