October 3rd
Today I learned the finish of the proof in the classification of finite abelian groups. We showed a few days ago that we can decompose any finite abelian group into a direct product of its Sylow $p$-subgroups, which left the case of $p$-groups. Then we proved the lemma that if a finite abelian $p$-group has a unique smallest subgroup, then it's cyclic.
We'll show that for $G$ a finite abelian $p$-group with cyclic subgroup $C$ of maximal order, we have that $G\cong C\times G'$ for some subgroup $G'\subseteq G,$ where the isomorphism is the canonical $(x,y)\mapsto xy.$ For closure, we show how to finish the proof assuming this. There is always a nontrivial cyclic subgroup for nontrivial $G,$ so applying this lemma repeatedly to $G$ and then $G'$ and then $G''$ and continuing down (until we're left with a trivial group) forms a progressively longer direct product\[G\cong C\times C'\times C''\times\cdots\]with progressively decreasing (in order) $G^{(\bullet)}$ remainder groups. But eventually the $G^{(\bullet)}$ will be trivial, completing the decomposition into a direct product of cyclic groups.
Once again, in order to not surprise the reader, we induct on $|G|.$ In the base case of $C=G,$ we can write $G\cong C\times\{e\},$ which is what we wanted. Note that this is forced in $|G|=p,$ which is cyclic generated by any nontrivial subgroup by Lagrange. So we may take $C\subsetneq G.$
As setup, $C\subsetneq G$ gives an element $h\in G\setminus C.$ Now, $h\ne e\in C$ lets us say that $\langle h\rangle$ has a subgroup of order $p$ (generated by $h^{\ord(h)/p}$), which we'll name $H.$ Our main player will be the canonical homomorphism\[G\to G/H.\]Unsurprisingly, we're going to use $\varphi$ to active our induction. Because $H$ is cyclic, we see that $C\cap H=\{e\},$ so $CH\cong C\times H,$ and $CH/H\cong(C\times H)/(\{e\}\times H)\cong C,$ so $CH/H$ is a pretty large cyclic subgroup of $G/H.$ How large? Well, the order of any element $gH$ in $G/H$ is no more than its order in $G$ is no more than $|C|,$ so $CH/H\cong C$ has maximal order in $G/H$ among cyclic subgroups.
So we trigger our induction and get to write $G/K\cong CH/H\times G_K'$ (with some hideous notation), for some subgroup $G_H'\subseteq G/K.$ Making this nicer to look at, we fix $G'$ to be the preimage of $G_H',$ and we naturally claim that $G\cong C\times G'.$ Note that $G'$ must contain the preimage of $e\in G_H',$ so $H\subseteq G',$ so we note that $G/H\cong CH/H\times G'/H$ implies\[G=CH\cdot G'H=CG'.\]But further, the fact we're working the canonical embedding $CH/H\times G'/H\to G/H$ lets us assert that $CH/H\cap G'/H=eH,$ so $CH\cap G'=\{e\}.$ So it follows from the above that $G=C\times G',$ which is what we wanted.