# Maximal subgroups of a finite p-group

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 $L$.

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Just to add to Geoff Robinson's answer below. A counterexample can be obtained by taking $p=2$, $G$ cyclic of order $4$, $U = \{1\}$ and $L$ the unique proper non-trivial subgroup of $G$. – user91132 May 25 '12 at 16:54
$G$ can not be cyclic by assumption. We can easily find counterexample by Geoff Robinson's answer. Let $G=\langle a \rangle \times \langle b \rangle$, where $|a|=4$. Then $a^2$ is in Frattini subgroup of $G$. Hence $L=\langle a^2 \rangle$ has no complement. – Wei Zhou May 26 '12 at 0:46

This is false in general. Consider the case (which you can reduce to by isomorphism theorems) that $U =1$ and $L$ then necessarily has order $p.$ You are asking for the existence of a complement to $L$ in $G$, for you would have $G = L \times M$ if there were such a maximal subgroup $M.$ There is such a subgroup $M$ if and only if $L$ is not contained in $\Phi(G),$ the Frattini subgroup of $G.$ More precisely, in general you are guaranteed to find the maximal subgroup you want if and only if $L/U$ is not contained in the Frattini subgroup of $G/U.$ Given a subgroup $U$, unless $G/U$ is elementary Abelian (ie an Abelian $p$-group of exponent $p )$, you can always find a subgroup $L$ containing $U$ with $[L:U] = p$ such that every maximal subgroup of $G$ containing $U$ will also contain $L.$
A s noted Prof. Robinson, this is false. However, this is true iff $L\not\le M\Phi(G)$. Indeed, let $\bar G=G/M\Phi(G)$; then $\bar L$ is a direct factor of $\bar G$ of order $p$. If $\bar G=\bar L\times\bar U$. Then $U$ is a maximal subgroup of $G$ and $U\cap L=M$.