1
$\begingroup$

I am currently studying the Krull-Schmidt theorem for groups. However, since I have only really every worked with finite groups before, I'm having trouble coming up with a nonexample for the Krull-Schmidt theorem. That is, is there a simple example of a group which is not isomorphic to a product of undecomposable subgroups? Thanks so much in advance!

$\endgroup$
  • 1
    $\begingroup$ $\mathbb{Q}/\mathbb{Z}$ is not isomorphic to a finite direct product of indecomposable subgroups. $\endgroup$ – Arturo Magidin Sep 30 '18 at 23:15
2
$\begingroup$

One canonical example would be the direct sum of $\Bbb{Z}$ with itself countably many times. (The countable direct product also works.) Arturo's example of $\Bbb{Q}/\Bbb{Z}$ also works...any partition of the primes gives $\Bbb{Q}/\Bbb{Z}$ as a direct product of the subgroups with only those primes in the denominators, but since there are infinitely many primes, if there are only finitely many factors one of them can always be further decomposed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.