# Help understanding proof of The Fundamental Theorem of Finite Abelian Groups.

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.

• He does strong induction, because he sais that assume inductively that it is tru for all groups whose order is less than $|H|$. So if you take $K$ such as $H=<a> \bigoplus K$ then $|K|<|H|$, so you can use the inductive hypothesis. By the way i don't like that proof so much, i recommend you the one in the book "José F. Fernando, J. Manuel Gamboa; Estructuras algebraicas" Commented Jun 18, 2022 at 11:55
• Please do not rely on pictures of text.
– Shaun
Commented Jun 18, 2022 at 12:19
• Honestly, I do not understand why someone would downvote this question just for having used a screenshot. Could someone explain why it is never acceptable to use a screenshot? Seems a bit strange in my opinion.
– Logi
Commented Jun 18, 2022 at 12:40

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.

• Just to be clear: the inductive step is that if it is true for all p-groups of order less than $H$, then it is also true for $H$?
– Logi
Commented Jun 18, 2022 at 12:01
• @Logi Precisely! Commented Jun 18, 2022 at 12:58
• Great, thank you!
– Logi
Commented Jun 18, 2022 at 13:03