- #1
Pratibha
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How to find the composition series of a group of order 30
MHB Composition Series for Group Order 30
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To find the composition series of a group of order 30, it is essential to determine whether the group is abelian or not. For abelian groups, the Abelian structure theorem applies, while non-abelian groups require additional analysis. A group of order 30 has a normal subgroup of index 2, leading to a normal subgroup of order 15. By applying the Sylow theorems, one can identify normal Sylow 5 and 3 subgroups, resulting in a composition series: <1> ≤ L ≤ H ≤ G, with factors of orders 2, 3, and 5. The Jordan-Hölder theorem confirms that all composition series will have the same factors.
- #2
Fallen Angel
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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.
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
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what is statement of abelian structure theorem?
- #4
johng1
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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
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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
Hello!
In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way:
I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true?
And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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