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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

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 $L$?2012-05-25
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    It's false even with $G$ non-cyclic. Take $G = \langle x \rangle \oplus \langle y \rangle$ with $2x=4y=0$, $L=\langle x,2y \rangle$, $U = \langle x \rangle$.2012-05-25
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    You need the condition $G/U$ *not* cyclic.2012-05-25

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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 cyclic2012-05-25
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    Try then a proof by some inductive argument on the number of different cyclic factors of G2012-05-25
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    Thanks for everyone. It is not true.2012-05-25
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    You're welcome. Perhaps you're interested in enhancing your acceptation rate?2012-05-25