17
$\begingroup$

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.)

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Greg Martin
    Commented May 13, 2014 at 1:59
  • $\begingroup$ 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). $\endgroup$
    – Benjamin Dickman
    Commented May 13, 2014 at 2:09
  • $\begingroup$ @BenjaminDickman : and the author is...?! $\endgroup$
    – DonAntonio
    Commented May 13, 2014 at 3:45
  • $\begingroup$ @DonAntonio Speiser, as can be found through google: scholar.google.com/… $\endgroup$
    – Benjamin Dickman
    Commented May 13, 2014 at 4:09
  • 2
    $\begingroup$ 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 $\endgroup$
    – Jack Schmidt
    Commented May 21, 2014 at 13:59

2 Answers 2

Reset to default
10
$\begingroup$

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.

enter image description here

The book The Theory of Finite Groups: An Introduction, by Kurzweil and Stellmacher, does what I had in mind in Chapter 2.

$\endgroup$
2
  • 1
    $\begingroup$ But what is "the lemma"? In particular, does it apply when $d=1$ (and thus $g=0$)? $\endgroup$
    – mr_e_man
    Commented Feb 13, 2020 at 1:55
  • $\begingroup$ Well, suppose we know that any vector $a\in\mathbb Z^n$ with GCD$(a)=1$ can be completed into a unimodular matrix $A$. Call its inverse $B$. Then we can define $$g_j=\sum_iA_{ij}x_i$$ (so $g=g_1$, after renaming $-a_1/d$ as $a_1$ etc.); it follows that $$\sum_iB_{ij}g_i=\sum_iB_{ij}\sum_kA_{ki}x_k$$ $$=\sum_k\sum_iA_{ki}B_{ij}x_k=\sum_k\delta_{kj}x_k=x_j.$$ This shows that anything generated by $\{x_1,\cdots,x_n\}$ is also generated by $\{g_1,\cdots,g_n\}$. $\endgroup$
    – mr_e_man
    Commented Feb 13, 2020 at 3:09
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Since the direct product of self-injective rings is self injective, this extends to the general case. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 23, 2014 at 1:51
  • $\begingroup$ If $H$ is a maximal cyclic subgroup of $G$ of order $n$, you can say that $G$ is a module over $\Bbb Z/n\Bbb Z$? $\endgroup$
    – Fabio Lucchini
    Commented Jun 2, 2018 at 22:27
  • $\begingroup$ @Fabio: this requires a lemma asserting that if an abelian group $A$ has elements of order $a, b$ then it has an element of order $\text{lcm}(a, b)$. It follows that if $n$ is the maximal order of an element of a finite $A$ then every element has order dividing $n$. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 30 at 19:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .