Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a test for a (finite) group in order to know if it is indecomposable? if so is this test equivelent to the definition of the indecomposable group

share|improve this question
    
Cross post from MO: mathoverflow.net/questions/82034/… –  Jack Schmidt Nov 30 '11 at 2:06
    
yes,it's also my post .what's the problem? –  user20216 Nov 30 '11 at 2:18
    
@user20216: It is considered bad manners to post in both fora without alerting users that the question has also been asked elsewhere. By alerting users that the question was also posted in mathoverflow, you inform users here so they can go check the post there and make sure they are not duplicating the effort of others (i.e., wasting their time). Symmetrically, you should inform people there that the post was made here for the very same reason. –  Arturo Magidin Nov 30 '11 at 4:45

1 Answer 1

I have no idea what your second question means. How can a test be equivalent to a definition?

I think the answer to the first question is that there is no straightforward way to do this and also no efficient algorithm to do it in general.

If you gave me a finite group and asked me whether it was indecomposable, and the answer was not immediately clear, then the first thing I would try would be to put it on a computer, find all of its normal subgroups, and check whether two of them complemented each other.

That would fail if there were too many normal subgroups. In that case I would try looking at the minimal normal subgroups. If the group is decomposable then there must be more than one of these, and the socle of the group (i.e. the group generated by its minimal normal subgroups) must itself be decomposable, so I would look at decompositions of that and see if they extended.

Do you have any specific examples, of types of examples in mind?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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