# Simple proof of the structure theorems for finite abelian groups

Many proofs of the structure theorems for finite abelian groups first reduce to the problem to $p$-groups, which is fine and is an important technique.

However, it seems to me that a simple proof can be based on this one crucial result:

Every finite abelian group has a cyclic direct factor. $\qquad(\star)$

From this, it follows at once by induction on the order of the group that

Every finite abelian group is a direct product of cyclic groups.

The Chinese remainder theorem then allows us to split these cyclic groups into cyclic $p$-groups and so we get the primary decomposition theorem:

Every finite abelian group is a direct product of cyclic $p$-groups.

The Chinese remainder theorem then allows us to reassemble these into an invariant factor decomposition.

Note that the proof is really easy once we have the crucial result $\star$ (and the Chinese remainder theorem of course).

I think Mac Lane and Birkoff in their Algebra do something along this line but it is not very simple.

Does anyone know a proof of the structure theorems along the lines above?

In particular, is there a simple proof of the crucial result $\star$ ?

(A side bonus would be a simple characterization of the cyclic subgroups that are direct factors of a given finite abelian group: I expect they are exactly the maximal cyclic subgroups.)

• Re: your last comment: there can be lots of cyclic direct factors, more than will fit in a single representation. For example, the group $Z_{12}\times Z_2$ can also be written as $Z_6\times Z_4$ and $Z_4\times Z_3\times Z_2$; so it has at least $5$ cyclic direct factors. – Greg Martin May 13 '14 at 1:59
• I don't see a copy online (and... it's in German) but you may wish to seek out: Die Theorie der Gruppen vonendlicher Ordnung (possibly volume 2). – Benjamin Dickman May 13 '14 at 2:09
• @BenjaminDickman : and the author is...?! – DonAntonio May 13 '14 at 3:45
• @DonAntonio Speiser, as can be found through google: scholar.google.com/… – Benjamin Dickman May 13 '14 at 4:09
• I would use Prüfer/Kulikoff's idea: a finite subgroup of an abelian group is a direct factor if and only if it is a pure subgroup. This is done in Kaplansky's short textbook on Infinite Abelian Groups - en.wikipedia.org/wiki/Pure_subgroup – Jack Schmidt May 21 '14 at 13:59

## 2 Answers

Rotman's Introduction to the Theory of Groups (theorem 6.9 on page 130) contains such a proof, attributed to Schenkman, but this proof is not exactly as I have outlined in the question. The book The Theory of Finite Groups: An Introduction, by Kurzweil and Stellmacher, does what I had in mind in Chapter 2.

For finite abelian $p$-groups you can do the following to split off a maximal cyclic subgroup as a direct summand.

If $q=p^n$ is the order of a maximal cyclic subgroup $H$ of our group $G$, then $G$ is a module over the ring $R=Z/qZ$. Now $H$ is free as an $R$-module, and $R$ is a self injective ring, so that $H$ is in fact injective. The inclusion $H\to G$ is therefore split.

• Since the direct product of self-injective rings is self injective, this extends to the general case. – Mariano Suárez-Álvarez May 23 '14 at 1:51
• If $H$ is a maximal cyclic subgroup of $G$ of order $n$, you can say that $G$ is a module over $\Bbb Z/n\Bbb Z$? – Fabio Lucchini Jun 2 '18 at 22:27