# Prove abelian non-cyclic p-group of order p^m (m>2) has a non-cyclic proper subgroup without structure theorem

It should have a copy of $\Bbb Z/p\Bbb Z\times \Bbb Z/p\Bbb Z$ but my brother's class has not learned it yet and this is for a homework problem... Any help?

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Given the additional condition, my answer fails to be applicable. – Cameron Buie Nov 5 '12 at 2:55

Let $G$ be our noncyclic, abelian $p$-group of order $p^m$. By Cauchy's theorem there is an element $x$ with $|x|=p$ which generates $X\subset G$. We construct another element $y$ of order $p$ outside $X$ as follows.
Take $y$ to be any element of $G\setminus X$ of minimal order. Then, $y^p\in X$. Either it is $1$ as desired, or it is $x^r$ for some integer $r\not\equiv p\pmod{p}$. It follows that the exponent $r$ can be inverted modulo $p$, so there is some $z\in X$ such that $z^p=y^p$.
Note that $z^{-1}y$ has order $p$ (since $G$ is abelian), and is not in $X$. Thus, $y$ has order $p$ by hypothesis.
We are done since the subgroup $X\times Y$, where $Y=\langle y\rangle$, is noncyclic of order $p^2<p^m$.