Composition Series for Group Order 30

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Discussion Overview

The discussion revolves around finding the composition series of a group of order 30, exploring both abelian and non-abelian cases, as well as the implications of the Sylow theorems and the structure theorem for abelian groups.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants suggest that the characteristics of the group, such as whether it is abelian, significantly influence the approach to finding the composition series.
  • One participant mentions the Abelian structure theorem, implying its relevance for abelian groups but does not provide details.
  • A participant proposes that a group of order 30 has a normal subgroup of index 2, leading to a specific composition series involving normal subgroups of orders 15, 5, and 3.
  • Another participant states that every finitely generated abelian group can be expressed as a direct product of cyclic groups, hinting at a structured approach to understanding abelian groups.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the Abelian structure theorem and the specifics of the composition series for groups of order 30, indicating that multiple competing views remain without a clear consensus.

Contextual Notes

The discussion does not resolve the assumptions regarding the nature of the group (abelian vs. non-abelian) or the completeness of the proposed composition series, leaving some mathematical steps and definitions unresolved.

Pratibha
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How to find the composition series of a group of order 30
 
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Hi,

It may depend on the caharacteristics of the group.

If your group it's abelian then use the Abelian's groups structure theorem.

If not, I think we need to know something more.
 
what is statement of abelian structure theorem?:confused:
 
This may be beyond your knowledge, but here goes:
A group of order 30 has a normal subgroup of index 2 (Any group of even order with a cyclic Sylow 2 subgroup has a normal subgroup of index 2; proved by considering the Cayley representation of G as a permutation group.) So let H be normal of order 15. By the Sylow theorems H has a normal Sylow 5 subgroup K (cyclic) and a normal Sylow 3 subgoup L. So one composition series (actually a chief series--all subgroups in series are normal in G) is:
$$<1>\,\trianglelefteq L\trianglelefteq H\trianglelefteq G$$ with the factors cyclic of orders 2, 3 and 5. The Jordan-Holder theorem says all composition series have the same factors.
 
Every finitely generated abelian group $G$ is isomorphic to a direct product

$\Bbb{Z}^{n}\oplus \Bbb{Z}_{a_{1}}\oplus \ldots \oplus \Bbb{Z}_{a_{k}}$

Moreover, we can order it in such a way that $a_{i}$ divides $a_{i+1}$ for every $i=1,\ldots, k-1$

PS: It's easier than I thought using the way johng has just posted
 

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