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.

A group $G$ is indecomposable if: $G = H \times K \Rightarrow \{ H,K \} = \{1, G \}$.
Then, a finite group $G$ decomposes into a direct product of indecomposable groups: $G = \prod_i G_i$.

Question: Is this decomposition unique (up to permutation and isomorphism)?

share|improve this question
1  
What does $\;\{H,K\}=\{1,G\}\;$ mean? –  Timbuc Aug 24 at 19:58
    
@Timbuc: it's an equality of sets. This means "$H=1$ and $K=G$" or "$H=G$ and $K=1$" (with $1$ the trivial group). –  Sébastien Palcoux Aug 24 at 19:59
    
@SébastienPalcoux What does $[H,K]$ mean, then? –  Thomas Andrews Aug 24 at 20:00
2  
@IttayWeiss, doesn't that theorem apply only to abelian groups? –  Timbuc Aug 24 at 20:03
1  
@Timbuc you are right. I was sure OP was talking about abelian groups. –  Ittay Weiss Aug 24 at 20:11

2 Answers 2

up vote 7 down vote accepted

Yes, the decomposition is unique for finite groups. This is a consequence of the Remak-Krull-Schmidt Theorem, which applies to groups that satisfy both the minimum and maximum conditions on normal subgroups. Since finite groups certainly have these properties, R-K-S applies here.

share|improve this answer

This decomposition is not unique in the class of all groups, although, as the other answer points out, it is true fir finite groups.

One reason that the result fails in general is that groups are not cancellable in general. A group $H$ is cancellable if the following holds. $$H\times Q\cong H\times P\Rightarrow P\cong Q$$ Finite groups are cancellable, but in general groups are not (see this question - $\mathbb{Z}$ is a counter-example). If your result held then the decompositions of $P$ and $Q$ would have to be the same, and hence $P$ and $Q$ they would have to be isomorphic, so all groups would be cancellable.

As finite groups are cancellable, this proof only works for general groups and not for finite groups.

share|improve this answer
1  
Your answer is a long comment, because my question is precisely for finite groups. Thank you for the link! –  Sébastien Palcoux Aug 24 at 21:21
2  
You start your answer by "No, this decomposition is not unique" and you finish by "but I suspect that it still does not hold". It would be more correct to start by "I suspect this decomposition is not unique" or "This decomposition is not unique in general and I suspect the same for the finite groups". –  Sébastien Palcoux Aug 24 at 21:22
1  
@Sebastian Sorry, my mistake, I didn't see the the word "finite". I've edited my answer to fit it in. –  user1729 Aug 25 at 8:12

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.