MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $L.$

share|cite|improve this answer

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 $U\cap L=M$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.