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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: enter image description here

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.

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    $\begingroup$ 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" $\endgroup$ Jun 18 at 11:55
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    $\begingroup$ Please do not rely on pictures of text. $\endgroup$
    – Shaun
    Jun 18 at 12:19
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    $\begingroup$ 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. $\endgroup$
    – Logi
    Jun 18 at 12:40

1 Answer 1

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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.

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  • $\begingroup$ 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 $H$? $\endgroup$
    – Logi
    Jun 18 at 12:01
  • $\begingroup$ @Logi Precisely! $\endgroup$
    – Gauss
    Jun 18 at 12:58
  • $\begingroup$ Great, thank you! $\endgroup$
    – Logi
    Jun 18 at 13:03

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