/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Let \(p, q\) be distinct primes.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 28

Let \(p, q\) be distinct primes. Prove that a group of order \(p^{2} q\) is solvable, and that one of its Svlow subgroups is normal.

Short Answer

Expert verified
In all possible cases, we have shown that a group G of order \(p^2 q\) must be solvable, which means it has a series of normal subgroups such that each factor group is abelian. Furthermore, one of its Sylow subgroups is always normal. We used the Sylow Theorems along with proof by contradiction to establish these results.

Step by step solution

01

Determine Sylow subgroups

Denote \(G_p\) and \(G_q\) as Sylow p-subgroups and Sylow q-subgroups of G, respectively. We have \(|G_p|=p^2\) and \(|G_q| = q\). We know that the number of Sylow p-subgroups, \(n_p\), divides q, and \(n_p \equiv 1 \pmod{p}\). Likewise, the number of Sylow q-subgroups, \(n_q\), divides \(p^2\), and \(n_q \equiv 1 \pmod{q}\).
02

Consider case where both Sylow subgroups are normal

Now, if either \(n_p = 1\) or \(n_q = 1\), then their respective Sylow subgroups are unique and therefore normal. In this case, let's assume we have both \(n_p = 1\) and \(n_q = 1\), meaning that G has a unique Sylow p-subgroup and a unique Sylow q-subgroup. Then, the internal direct product \(G_p \times G_q\) is also a subgroup of G with order \(p^2 \times q = |G|\), and G must be the internal direct product of its Sylow subgroups. Since both G_p and G_q are abelian, being prime power order, G would be solvable and have a normal Sylow subgroup.
03

Consider case where one Sylow subgroup is normal

In this case, assume without loss of generality that \(n_p = 1\) so that G has a unique Sylow p-subgroup G_p. Then, G_p is normal in G, and we can form the factor group \(G/G_p\), with order q, which is also abelian. Now we use the result that if H is normal in G and both H and G/H are solvable, then G is solvable. Since \(G_p\) and \(G/G_p\) are abelian, they are both solvable. Then G must be solvable as well.
04

Consider case where none of the Sylow subgroups are normal

In the last case, where neither of the Sylow subgroups is normal, we would have \(n_p = q\) and \(n_q = p^2\) or \(n_q = p\). But \(n_p \equiv 1 \pmod{p}\), so in this case, we must have \(n_q = p\), as \(n_q=p^2\) is not congruent to 1 modulo q. Since there are p Sylow q-subgroups, each with q-1 nonidentity elements, and their intersection is trivial, we have a total of \(p\times(q-1)\) elements in G that are not in the p-subgroups. However, this would mean that there are only \(p^2 q - p(q-1) = p^2\) elements left in G, but we need at least \(p^2 q\) elements for the union of the Sylow p-subgroups. This contradiction implies that our initial assumption is incorrect - one of the Sylow subgroups must be normal.
05

Conclusion

In all possible cases, we have shown that the group G of order \(p^2 q\) is solvable and one of its Sylow subgroups must be normal, as requested.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 focuses on studying algebraic structures known as groups. A group is a collection of elements with a binary operation that combines any two elements to form a third element, while satisfying four essential conditions: closure, associativity, identity, and invertibility.

For instance, when we talk about a group of order \(p^2 q\), where \(p\) and \(q\) are distinct prime numbers, we are referring to a group that has \(p^2 q\) elements. In the context of Sylow subgroups, which are subsets of a group that have a size equal to a power of a prime number and satisfy certain divisibility properties, Group Theory provides tools to analyze the structure and properties of the group.
Solvable Group
A solvable group is a type of group that can be broken down into a series of subgroups, in which each subgroup is normal in the next one, and the factor groups of consecutive subgroups are abelian. Essentially, solvable groups can be decomposed into simpler pieces that are easy to understand and work with.

Analogous to solving a complex puzzle by breaking it into smaller, manageable parts, identifying a group as solvable allows mathematicians to gradually unravel its structure. In the exercise, when the group of order \(p^2 q\) is deemed solvable, it is an affirmation that the group can be 'solved' or understood in a step-by-step manner by examining its constituent abelian subgroups.
Normal Subgroup
A normal subgroup of a group is a subset that is invariant under group conjugation by elements from the entire group. This means that the subgroup remains unchanged when any of its elements is 'conjugated' by any element from the main group.

The significance of a subgroup being normal is substantial. Not only does it enable the construction of factor groups which are crucial for analyzing the structure of the original group, but it also serves as an indicator of the group's internal symmetry. The exercise highlights the presence of a normal Sylow subgroup as a key criterion in determining the solvability of the group.
Prime Order Groups
Groups of prime order hold a special spot in Group Theory due to their simplicity and well-understood structure. A prime order group is a group with a number of elements equal to a prime number. These groups are always cyclic and, by definition, abelian, which means that every element commutes with every other element.

In the provided solution, the group of order \(q\), which is a prime number, and the Sylow q-subgroups of the group of order \(p^2 q\), are examples of prime order groups. Their simplicity plays a vital role in determining whether the larger group is solvable, as abelian groups are by default solvable, further contributing to the overall solvability of the group.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(Tits) Let \(M_{1}, \ldots, M_{r} \in G L_{2}(C)\) be a finite number of matrices. Let \(\lambda_{i}, \lambda_{i}^{\prime}\) be the eigenvalues of \(M_{i}\). Assume that each \(M_{i}\) has two distinct complex fixed points, and that \(\left|\lambda_{i}\right|<\left|\lambda_{i}^{\prime}\right|\). Also assume that the fixed points for \(M_{1}, \ldots, M\), are all distinct from each other. Prove that there exists a positive integer \(k\) such that \(M_{1}^{k}, \ldots, M_{r}^{k}\) are the free generators of a free subgroup of \(G L_{2}(C) .\left[\right.\) Hint: Let \(w_{i}, w_{i}^{\prime}\) be the fixed points of \(M_{i} .\) Let \(U_{i}\) be a small disc centered at \(w_{i}\) and \(U_{i}^{\prime}\) a small disc centered at \(w_{i}^{\prime}\). Let \(S_{i}=U_{i} \cup U_{i}^{\prime}\). Let \(s\) be a complex number which does not lie in any \(S_{i}\). Let \(G_{i}=\left(M_{i}^{k}\right) .\) Show that the conditions of Exercise 54 are satisfied for \(k\) sufficiently large.].

Let \(G\) operate on a set \(S\). Let \(S=\bigcup S_{i}\) be a partition of \(S\) into disjoint subsets. We say that the partition is stable under \(G\) if \(G\) maps each \(S_{i}\) onto \(S_{j}\) for some \(j\), and hence \(G\) induces a permutation of the sets of the partition among themselves. There are two partitions of \(S\) which are obviously stable: the partition consisting of \(S\) itself, and the partition consisting of the subsets with one element. Assume that \(G\) operates transitively, and that \(S\) has more than one element. Prove that the following two conditions are cquivalent: PRIM 11\. The only partitions of \(S\) which are stable are the two partitions mentioned above. PRIM 2. If \(H\) is the isotropy group of an element of \(S\), then \(H\) is a maximal subgroup of \(G\). These two conditions define what is known as a primitive group, or more accurately, a primitive operation of \(G\) on \(S\). Instead of saying that the operation of a group \(G\) is 2-transitive, one also says that it is doubly transitive.

(H K)=\frac{\\#(H) \\#(K)}{\\#(H \cap K)} $$ # Let \(H, K\) be subgroups of a finite group \(G\) with \(K \subset N_{H}\). Show that $$ \\#(H K)=\frac{\\#(H) \\#(K)}{\\#(H \cap K)} $$

Let \(G\) be a group acting transitively on a set \(S\) with at least 2 elements. For each \(x \in G\) let \(f(x)=\) number of elements of \(S\) fixed by \(x\). Prove: (a) \(\sum_{x \in G} f(x)=\\#(G)\). (b) \(G\) is doubly transitive if and only if $$ \sum_{x \in G} f(x)^{2}=2 \\#(G) $$

Let \(\sigma=[123 \cdots n]\) in \(S_{n}\). Show that the conjugacy class of \(\sigma\) has \((n-1) !\) elements. Show that the centralizer of \(\sigma\) is the cyclic group generated by \(\sigma .\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.