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Chapter 13: Problem 2

Find all of the abelian groups of order 200 up to isomorphism.

Short Answer

Expert verified
Question: List all abelian groups of order 200 up to isomorphism. Answer: The abelian groups of order 200 up to isomorphism are: 1. \(\mathbb{Z}_{200}\) 2. \(\mathbb{Z}_8 \oplus \mathbb{Z}_{25}\) 3. \(\mathbb{Z}_8 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5\) 4. \(\mathbb{Z}_4 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{25}\) 5. \(\mathbb{Z}_4 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5\) 6. \(\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{25}\) 7. \(\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5\)

Step by step solution

01

Find the prime factorization of 200

Find the prime factors of 200: \(200 = 2^3 \cdot 5^2\).
02

Determine possible abelian groups

Using the Fundamental Theorem of Finite Abelian Groups, we can determine the possible abelian groups of order 200 by considering the possible direct sums of cyclic groups for each prime factor. For the \(2^3\) part, we have three possibilities: 1. One cyclic group of order \(2^3 = 8\) 2. A direct sum of three cyclic groups of order \(2^1 = 2\) (because \(2 \cdot 2 \cdot 2 = 8\)) 3. A direct sum of one cyclic group of order \(2^2 = 4\) and another of order \(2^1 = 2\) (because \(4 \cdot 2 = 8\)) For the \(5^2\) part, we have two possibilities: 1. One cyclic group of order \(5^2 = 25\) 2. A direct sum of two cyclic groups of order \(5^1 = 5\) (because \(5 \cdot 5 = 25\)) We will now combine these possibilities:
03

Combine the possibilities for both prime factors

By taking the direct sums of the possibilities for both prime factors, we will obtain the possible abelian groups of order 200. 1. One cyclic group of order \(2^3 \cdot 5^2 = 200\), which gives the abelian group \(\mathbb{Z}_{200}\). 2. A direct sum of one cyclic group of order \(2^3 = 8\) and another of order \(5^2 = 25\), or the group \(\mathbb{Z}_8 \oplus \mathbb{Z}_{25}\). 3. A direct sum of one cyclic group of order \(2^3 = 8\) and the direct sum of two cyclic groups of order \(5^1 = 5\), or the group \(\mathbb{Z}_8 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5\). 4. A direct sum of one cyclic group of order \(2^2 = 4\), another of order \(2^1 = 2\), and one of order \(5^2 = 25\), or the group \(\mathbb{Z}_4 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{25}\). 5. A direct sum of one cyclic group of order \(2^2 = 4\), another of order \(2^1 = 2\), and the direct sum of two cyclic groups of order \(5^1 = 5\), or the group \(\mathbb{Z}_4 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5\). 6. A direct sum of three cyclic groups of order \(2^1 = 2\) and one of order \(5^2 = 25\), or the group \(\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{25}\). 7. A direct sum of three cyclic groups of order \(2^1 = 2\) and the direct sum of two cyclic groups of order \(5^1 = 5\), or the group \(\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5\). Thus, there are seven abelian groups of order 200 up to isomorphism.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Theorem of Finite Abelian Groups
Understanding the structure of finite abelian groups is made simpler by the Fundamental Theorem of Finite Abelian Groups. This theorem states that any finite abelian group can be expressed as a direct sum of cyclic subgroups of prime power orders. This means if you have an abelian group of a certain order, you can break it down into simpler, more manageable components, specifically cyclic groups whose orders are powers of prime numbers.
  • To use this theorem, start by factorizing the order of the group into its prime components.
  • Then, consider the different ways to organize these prime powers into cyclic groups and their direct sums.
  • Each unique way provides a different abelian group up to isomorphism.
This theorem is particularly useful because it gives us a framework to list all possible abelian groups of a given order by considering combinations of direct sums and cyclic groups.
Cyclic Groups
Cyclic groups are among the simplest types of groups in abstract algebra. A cyclic group is generated by a single element, meaning every element in the group can be expressed as a power of this generator. If you think of it visually, it's like wrapping a string around a circle, where each overlapping signifies a repeat.
To determine if a group is cyclic:
  • Check if there is an element (generator) such that every group element can be created by repeatedly applying the group operation to this element.
  • A group of prime order is always cyclic.
Cyclic groups can be finite or infinite, but importantly, those involved in abelian group structures, especially under the Fundamental Theorem, are always finite cyclic groups.
Direct Sum
The concept of the direct sum is essential when discussing abelian groups because it allows us to combine smaller groups into a single larger one in a straightforward manner. For abelian groups composed using the Fundamental Theorem, the direct sum operation provides a way to represent these groups as combinations of simpler cyclic groups.
Here's how it works:
  • Each group in the direct sum operates independently of the others.
  • The order of the entire direct sum group is the product of the orders of its components.
For example, if you have two groups, one of order 4 and one of order 5, their direct sum is an abelian group of order 20. This highlights the significance of the direct sum in constructing and analyzing the structure of finite abelian groups.
Group Isomorphism
Group isomorphism is a critical concept that implies two groups are structurally identical even if their elements and operations appear different. When we say two groups are isomorphic, we mean there exists a one-to-one correspondence between their elements that preserves group operations.
Key points about group isomorphism include:
  • Isomorphic groups have the same structure, which means they are essentially the same group, distinguished only by the labeling of elements.
  • In the context of abelian groups, identifying isomorphisms helps classify groups into distinct types rather than counting them as separate when they have the same structure.
By demonstrating an isomorphism between two groups, we affirm they share the same fundamental properties, making this a powerful tool in group theory for simplifying and solving problems.

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Most popular questions from this chapter

Show that the infinite direct product \(G=\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \cdots\) is not finitely generated.

Find all of the composition series for each of the following groups. (a) \(\mathbb{Z}_{12}\) (b) \(\mathbb{Z}_{48}\) (c) The quaternions, \(Q_{8}\) (d) \(D_{4}\) (e) \(S_{3} \times \mathbb{Z}_{4}\) (f) \(S_{4}\) \((\mathrm{g}) S_{n}, n \geq 5\) (h) \(\mathbb{Q}\)

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Zassenhaus Lemma. \(\quad\) Let \(H\) and \(K\) be subgroups of a group \(G\). Suppose also that \(H^{*}\) and \(K^{*}\) are normal subgroups of \(H\) and \(K\) respectively. Then (a) \(H^{*}\left(H \cap K^{*}\right)\) is a normal subgroup of \(H^{*}(H \cap K)\). (b) \(K^{*}\left(H^{*} \cap K\right)\) is a normal subgroup of \(K^{*}(H \cap K)\). (c) \(H^{*}(H \cap K) / H^{*}\left(H \cap K^{*}\right) \cong K^{*}(H \cap K) / K^{*}\left(H^{*} \cap K\right) \cong(H \cap K) /\left(H^{*} \cap K\right)\left(H \cap K^{*}\right)\)
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