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Chapter 15: Problem 6

Prove that no group of order 160 is simple.

Short Answer

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Question: Prove that no group of order 160 is simple. Solution: We find the prime factorization of the group order (160), which is \(2^5 * 5\). We consider the possible Sylow subgroups for each of the prime factors. For the prime factor 2, we have Sylow 2-subgroups of order 32. Analyzing the possible number of Sylow 2-subgroups, n_2, we find it can only be 1. Therefore, there must always be a normal Sylow 2-subgroup in a group of order 160. Since a simple group has no proper nontrivial normal subgroups, no group of order 160 can be simple.

Step by step solution

01

Finding Prime Factors of the Group Order

Find the prime factorization of the group order (160). The prime factorization is \(2^5 * 5\).
02

Sylow Theorems

The Sylow theorems help us determine the possibilities for the number of Sylow subgroups with a given prime power order. For each of the prime factors found in Step 1, we need to apply the Sylow theorems to understand the structure of the potential subgroups.
03

Sylow p-Subgroups for 2

For the prime factor 2, we have \({2^5}\) which gives us Sylow 2-subgroups of order 32. According to Sylow's third theorem, the number of Sylow 2-subgroups (denoted as n_2) must divide the index, which in this case is 5, and must be equal to 1 modulo 2. Therefore, n_2 can be 1 or 5.
04

Sylow p-Subgroups for 5

For the prime factor 5, we have \({5^1}\) which gives us Sylow 5-subgroups of order 5. Again, according to Sylow's third theorem, the number of Sylow 5-subgroups (denoted as n_5) must divide the index, which in this case is 32, and must be equal to 1 modulo 5. Thus, n_5 can be 1, 2, 4, 8, 16, or 32.
05

Normal Subgroups

Since n_2 can be either 1 or 5, we analyze both cases. If n_2 = 1, there is only one Sylow 2-subgroup, which must be normal, as it is the only subgroup of its order. So the group is not simple. If n_2 = 5, we must have 5 Sylow 2-subgroups. This means that there are \(5*(32-1) = 155\) elements of order 2. Since we have 160 elements in total, there are only 5 remaining elements in the group. However, we know that there must be at least one Sylow 5-subgroup of order 5, so there is a contradiction. This means n_2 cannot be equal to 5, and n_2 must be 1. In conclusion, there must always be a normal Sylow 2-subgroup in a group of order 160. Therefore, no group of order 160 can be simple.

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Key Concepts

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

Group Theory
Group Theory is a branch of mathematics that studies the algebraic structures known as groups. A group consists of a set equipped with an operation that combines any two elements to form a third element, also within the set, and this operation satisfies four fundamental properties:
  • Closure: For all elements \(a\) and \(b\) in the group, the result of the operation, \(a\cdot b\), is also in the group.
  • Associativity: The group operation is associative; that means \((a \cdot b) \cdot c = a \cdot (b \cdot c)\) for any elements \(a, b, c\).
  • Identity Element: There exists an identity element \(e\) such that for every element \(a\), the equation \(e \cdot a = a \cdot e = a\) holds true.
  • Inverse Element: For each element \(a\), there exists an element \(b\) such that \(a \cdot b = b \cdot a = e\).
Group Theory is fundamental in various mathematical areas and real-world applications such as cryptography and physics. Understanding groups helps in analyzing symmetrical structures and their properties.
Sylow p-subgroups
Sylow p-subgroups are a key concept in Group Theory, named after the Norwegian mathematician Ludwig Sylow. These subgroups play a crucial role in understanding the structure of groups whose order (the number of elements) is divisible by a power of a prime number \(p\).
  • A Sylow p-subgroup of a group \(G\) is a maximal p-subgroup, meaning it is a subgroup of \(G\) with an order of \(p^n\), where \(p^n\) divides the order of \(G\), but \(p^{n+1}\) does not.

  • The Sylow Theorems provide information about these subgroups, giving conditions for their existence and uniqueness, and constraints on the number of such subgroups.
  • The third Sylow theorem states that the number of Sylow p-subgroups, denoted by \(n_p\), divides the order of the group and is congruent to 1 modulo \(p\).
Using these theorems allows us to deduce important properties of a group and prove whether a group contains a normal Sylow subgroup, thus showing that it's not simple.
Normal Subgroups
In group theory, a normal subgroup is a subgroup that is invariant under conjugation by any element of the group. This means that if \(N\) is a normal subgroup of \(G\), then for every element \(n\) in \(N\) and every \(g\) in \(G\), the element \(gng^{-1}\) is also in \(N\).
  • Normal subgroups are important because they allow us to construct quotient groups, which can simplify complex group structures.

  • A group is simple if it has no normal subgroups other than the trivial subgroup and the group itself.
  • In the context of the exercise, if a group of order 160 has a non-trivial normal subgroup, it cannot be simple.
When analyzing the number of Sylow subgroups, if either has \(n_p = 1\), this implies the presence of a normal subgroup, which is crucial in determining the simplicity of the group. A group where a Sylow p-subgroup is normal indicates the group's membership structure isn't simple, reinforcing the conclusion of the exercise.

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Most popular questions from this chapter

Let \(G\) be a group of order \(p^{r}, p\) prime. Prove that \(G\) contains a normal subgroup of order \(p^{r-1}\).

Let \(H\) be a Sylow \(p\) -subgroup of \(G\). Prove that \(H\) is the only Sylow \(p\) -subgroup of \(G\) contained in \(N(H)\).

The Frattini Lemma. If \(H\) is a normal subgroup of a finite group \(G\) and \(P\) is a Sylow \(p\) -subgroup of \(H,\) for each \(g \in G\) show that there is an \(h\) in \(H\) such that \(g P g^{-1}=h P h^{-1}\) Also, show that if \(N\) is the normalizer of \(P,\) then \(G=H N\).

Show that if the order of \(G\) is \(p^{n} q,\) where \(p\) and \(q\) are primes and \(p>q,\) then \(G\) contains a normal subgroup.

(a) Suppose \(p\) is prime and \(p\) does not divide \(m\). Show that $$ p \nmid\left(\begin{array}{c} p^{k} m \\ p^{k} \end{array}\right) . $$ (b) Let \(\mathcal{S}\) denote the set of all \(p^{k}\) element subsets of \(G\). Show that \(p\) does not divide \(|\mathcal{S}|\). (c) Define an action of \(G\) on \(\mathcal{S}\) by left multiplication, \(a T=\\{\) at \(: t \in T\\}\) for \(a \in G\) and \(T \in \mathcal{S}\). Prove that this is a group action. (d) Prove \(p \nmid\left|\mathcal{O}_{T}\right|\) for some \(T \in \mathcal{S}\). (e) Let \(\left\\{T_{1}, \ldots, T_{u}\right\\}\) be an orbit such that \(p \nmid u\) and \(H=\left\\{g \in G: g T_{1}=T_{1}\right\\}\). Prove that \(H\) is a subgroup of \(G\) and show that \(|G|=u|H|\) (f) Show that \(p^{k}\) divides \(|H|\) and \(p^{k} \leq|H|\). (g) Show that \(|H|=\left|\mathcal{O}_{T}\right| \leq p^{k} ;\) conclude that therefore \(p^{k}=|H| .\)
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