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\title{Lemma for FTFAG}

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\begin{frame}\frametitle{Strategy (``Peel off'' a cyclic summand (for FTFAG))}

\bl {\bf Lemma:} Let $G$ be a finite abelian $p$-group, and let $a$ be an element of maximal order $p^m$ in $G.$  Then there is a subgroup $K$ of $G$ such that $G=<a>\oplus K.$
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\visible<2->{\item<beamer:check@8-> Induct on the order $p^n$ of $G.$}
\visible<3->{\item<beamer:check@9-> Choose an element $b\in G\setminus <a>$ of minimal order and use $b$ to construct an element $c$ of order $p$ in $G\setminus <a>.$}
\visible<4->{\item<beamer:check@10-> Conclude $|b|=p$ and thus $<a>\cap <b>=\{0\}.$}
\visible<5->{\item<beamer:check@11-> Show $a+<b>$ has maximal order in $G/<b>,$ which has order $p^{n-1}.$}
\visible<6->{\item<beamer:check@12-> Write $G/<b>=<a+<b>>\oplus \ov{K}$ for some $\ov{K}\le G/<b>.$  Let $K\le G$ be the ``pullback'' of $\ov{K}$.}
\visible<7->{\item<beamer:check@13-> Show $<a>\cap K=\{0\}$ and $G=<a>+K$ (counting argument).}
\ee



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