7
$\begingroup$

A first thing that we do to analyze something is "divide and conqur." So it is natural to consider simple groups. In the same way, it appears to be natural to consider indecomposable groups. However, no one taught me about indecomposable groups and I know it quite recently.

Of course, indecomposable groups is larger class than simple groups. Having said that, it doesn't answer my question fully because a lager concept is sometimes easier to deal with (for example, $\mathbb{Z} \subseteq \mathbb{R}$).

Is there any reason that we care about simple groups rather than indecomposable groups?

Added: From the comment to DustanLevenstein — For example, in module theory, indecomposable modules looks be treated equal with simple modules. importance of indecomposable modules and simple modules seems be at the same level. What's the difference between these?

$\endgroup$
  • 1
    $\begingroup$ Classifying all finite indecomposable groups is equivalent to classifying all finite groups (because the "extension problem" for indecomposable groups becomes rendered trivial, by definition). The fact that we have successfully classified finite simple groups but not all finite groups should give an indication that the latter is a much more difficult problem, even given the former. That's not really an answer to your question, just something that I hope will help catalyse your thoughts on this. $\endgroup$ – Dustan Levenstein Aug 9 '14 at 14:41
  • $\begingroup$ @DustanLevenstein Thank you for your comment. If motivation to consider (finite) simple groups is to consider (finite) groups, then isn't it natural to consider indecomposable groups since its extension problem is trivial? Besides, even if classifying those completely is hard, investigating some properties of indecomposable groups doesn't help to solve some problems? For example, in module theory, indecomposable modules looks be treated equal with simple modules. What's the difference between these? $\endgroup$ – Orat Aug 10 '14 at 5:44
  • $\begingroup$ @Taro Have you read this old question? It doesn't tell you "why not indecomposable groups", but rather tells you "why simple groups". However, the underlying reason for looking at simple groups (the Jordan-Hölder theorem, which is what the linked question is about) doesn't work for indecomposable groups. $\endgroup$ – user1729 Aug 12 '14 at 9:28
  • $\begingroup$ @user1729 Thank you for pointing out about the question. At least for Jordan-Holder theorem, there is a similar one: Krull-Remak-Schmidt theorem. $\endgroup$ – Orat Aug 12 '14 at 9:32
  • $\begingroup$ @Taro Ah, thank you. My comment was supposed to end with a "...so far as I know." Thanks for pointing that out! $\endgroup$ – user1729 Aug 12 '14 at 9:34
9
+50
$\begingroup$

Making good mathematical definitions involves contemplating a tradeoff between generality and power. For example, monoids are more general than groups, but one reason people usually don't learn about monoids in a first course on abstract algebra is that it's hard to prove anything substantive about even, say, finite monoids without further assumptions, whereas groups are less general but one can prove lots of interesting things about, say, finite groups (e.g. Lagrange's theorem, the Sylow theorems).

Indecomposable groups are certainly more general than simple groups, and it is true that if we understood indecomposable groups then understanding all groups would be straightforward, but the difficulty hasn't been magically removed. The point of Dustan Levenstein's comment is that because it's so easy to classify all finite groups once we've classified all finite indecomposable groups, the latter should be essentially as hard to understand as the former. In particular one should not be able to make much stronger claims about indecomposable groups than one would be able to make about arbitrary groups: it's a bad tradeoff of generality vs. power.

Module theory is closer to linear algebra than to (nonabelian) group theory and so one can hope for a closer relationship between indecomposability and simplicity in this case. I don't have more specific knowledge about what happens in this case though.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +1. Perhaps, "In particular one should not be able to make much stronger claims about indecomposable groups than one would be able to make about arbitrary groups" might be an answer to my question. But I'll wait a week to see whether another point of view appears or not. $\endgroup$ – Orat Aug 12 '14 at 9:53

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.