- #1
Pratibha
- 5
- 0
How to find the composition series of a group of order 30
Composition Series for Group Order 30
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SUMMARY
The discussion focuses on finding the composition series of a group of order 30. It establishes that if the group is abelian, the Abelian groups structure theorem applies. For non-abelian groups, it identifies a normal subgroup of index 2, leading to a composition series represented as <1> ≤ L ≤ H ≤ G, where H is of order 15, and the factors are cyclic groups of orders 2, 3, and 5. The Jordan-Hölder theorem confirms that all composition series share the same factors.
PREREQUISITES- Understanding of group theory concepts, specifically composition series
- Familiarity with the Abelian groups structure theorem
- Knowledge of Sylow theorems and their applications
- Basic understanding of the Jordan-Hölder theorem
- Study the Abelian groups structure theorem in detail
- Explore Sylow theorems and their implications for group structure
- Learn about the Jordan-Hölder theorem and its applications in group theory
- Investigate examples of composition series for various group orders
Mathematicians, particularly those specializing in group theory, educators teaching abstract algebra, and students seeking to deepen their understanding of composition series and group structures.
- #2
Fallen Angel
- 202
- 0
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.
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.
- #3
Pratibha
- 5
- 0
what is statement of abelian structure theorem?
- #4
johng1
- 234
- 0
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.
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.
- #5
Fallen Angel
- 202
- 0
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
$\Bbb{Z}^{n}\oplus \Bbb{Z}_{a_{1}}\oplus \ldots \oplus \Bbb{Z}_{a_{k}}$
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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