October 1st
Today I learned the next lemma in the classification of finite abelian groups. To review, we currently are able to decompose finite abelian groups into direct products of their Sylow $p$-subgroups, so it remains to decompose $p$-groups into a direct product of cyclic $p$-groups.
The lemma is that if a finite abelian $p$-group $G$ has a unique smallest subgroup $H,$ then $G$ is actually cyclic. So as not to surprise the reader, we say that we're going to (strongly) induct on $|G|.$ Our base case of $|G|=p$ holds because Lagrange says that each non-identity element generates $G.$ So we may assume $|G| \gt p.$
We start by putting a name on $H.$ We note that by Cauchy that there is a (cyclic) subgroup of minimal order $p,$ so $H$ must have this order, and in fact $H$ must be made up of all the elements of order $p,$ for these each generate a subgroup of order $p.$ More succinctly, if we define $\varphi:g\mapsto g^p,$ then\[\ker(\varphi)=\left\{g\in G:g^p=e\right\}.\]Now, $\ker(\phi)$ consists exactly of the elements of order $p,$ all of which generate $H,$ so $\{e\}\subsetneq\ker(\phi)\subseteq H,$ so $H$ being smallest requires $H=\ker(\varphi).$ Note that $\ker(\varphi)\subsetneq G.$
The advantage of having homomorphism $\varphi$ is that the homomorphism theorem says that $\varphi(G)\cong G/\ker(\varphi),$ and $G/\ker(\varphi)$ is a smaller group than $G,$ so the inductive hypothesis lets us extract a coset $g\ker(\varphi)$ for which\[\langle g\ker(\varphi)\rangle=G/\ker(\varphi)\]generating $G/K.$ So we claim that $g$ also generates $G.$ We already know that every element $h\in G$ is in some coset $g^\bullet\ker(\varphi),$ so $h=g^\bullet k$ for $k\in\ker(\varphi).$ So we only need to prove that $\ker(\varphi)\subseteq\langle g\rangle.$ But $\langle g\rangle$ is nontrivial ($\langle g\ker(\varphi)\rangle$ is nontrivial), so it has $H=\ker(\varphi)$ is a subgroup, which is what we wanted.