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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.

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Let $G=C_4$, $L=\langle 2\rangle$, $U=\langle 0\rangle$. What maximal subgroup of $G$ will you pick that does not contain $L$? – Arturo Magidin May 25 '12 at 13:37
It's false even with $G$ non-cyclic. Take $G = \langle x \rangle \oplus \langle y \rangle$ with $2x=4y=0$, $L=\langle x,2y \rangle$, $U = \langle x \rangle$. – Derek Holt May 25 '12 at 13:54
You need the condition $G/U$ not cyclic. – j.p. May 25 '12 at 13:58

Your proof fails if G is cyclic, by the uniqueness of sbgps. of a given divisor of the order of the group.

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assume G is not cyclic – user28083 May 25 '12 at 13:46
Try then a proof by some inductive argument on the number of different cyclic factors of G – DonAntonio May 25 '12 at 13:51
Thanks for everyone. It is not true. – user28083 May 25 '12 at 14:18
You're welcome. Perhaps you're interested in enhancing your acceptation rate? – DonAntonio May 25 '12 at 14:25

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