In Hungerford's Abstract Algebra: An Introduction, he writes The Fundamental Theorem of Finite Abelian Groups as:
"Every finite abelian group G is the direct sum of cyclic groups, each of prime power order.
The following is the proof for this theorem:
My problem is that I do not understand the inductive proof - where is the inductive step? The basis step is that the assertion is true when $H$ has order $2$, which was shown previously in the book, but it looks like they just take the induction step for granted.
When he takes an element $a$ of maximal order and writes $H = \langle a \rangle \oplus K$, since the order of $a$ is greater than $1$, the order of $K$ is less than the order of $H$. Moreover, $K$ is a $p$-group, because it is a subgroup of a $p$-group.
The induction hypothesis is “for every $p$-group of order less than $\mid H \mid$, that $p$-group is the direct sum of cyclic groups.”
This means that $K$ is a group that satisfies the induction hypothesis, and this is where it is used.
I also consider the proof a bit odd, and would probably fix an arbitrary prime $p$ beforehand, and do an induction on the exponent of that prime. The proof would then go by exactly the same.