Purchase Solution

Groups and Subgroups of G

Not what you're looking for?

Ask Custom Question

Let G sub1 and G sub 2 be groups, with subgroups H sub 1 and H sub 2, respectivetly. Show that {(x sub 1, x sub 2) | H sub 1 is an element of H sub 1, x sub 2 is an element of H sub 2} is a subgroup of the direct product G sub 1 x G sub 2.

Purchase this Solution
Solution Summary

Groups and Subgroups are investigated. The solution is detailed and well presented. The solution received a rating of "5" from the student who posted the question.

Purchase this Solution
Free BrainMass Quizzes
Exponential Expressions

In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.

Know Your Linear Equations

Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

Probability Quiz

Some questions on probability

View More Free Quizzes
Related BrainMass Solutions

  • Proof that There Are No Simple Groups of Order 56 or 148

    We know from Sylow's third theorem that the number of Sylow-2 subgroups of G is congruent to 1 modulo 2 and divides 7, whence there is are either one or seven Sylow-2 subgroups of G, We also know that the number of Sylow-7 subgroups of G is congruent

  • Odd Order and Cyclic Groups

    97559 Odd Order and Cyclic Groups Suppose that G is a finite group of odd order 2n + 1. Prove or disprove that the number of cyclic subgroups of G is at most n + 1. The statement is true.

  • Groups

    Show that G is not the internal direct product of any two of its proper subgroups. If G is the internal direct product of subgroups G_1 and G_2 show that G/G_1 is isomorphic to G_2 and G/G_2 is isomorphic to G_1.

  • Orders of Cyclic Groups, Prime Order and Subgroups

    43682 Orders of Cyclic Groups, Prime Order and Subgroups (1) Let G be cyclic of order pn, p prime. Let H,K < G. Show that either H C K or K C H. (2) Let G 6 ≠ {e} such that it has no proper subgroups. Then G must be cyclic of prime order.

  • Permutation Groups

    This solution is comprised of a detailed explanation to prove that there exists subgroups Ni ( for some r) such that G = N0 > N1 >N2 >N3 > ... >Nr = (e) where Ni is a normal subgroup of Ni-1 and where Ni-1 /Ni is abelian.

  • Simple Groups and Sylow p-Subgroups

    147201 Simple Groups and Sylow p-Subgroups (1) Prove that if |G|=1365, then G is not simple. (2) Assume that G is a nonabelian group of order 15. Prove that Z(G)=1.

  • Sylow P-Subgroups, Isomorphisms and Solvable Groups

    104755 Sylow P-Subgroups, Isomorphisms and Solvable Groups 1. a) If M and N are normal subgroups of G then G/M is isomorphic to a subgroup of G/M x G/N. b) If G/M and G/N are solvable, then G/(M intersect N) is solvable. 2.

  • intersection of normal subgroups is a normal subgroup

    Indeed, if a<-K, for any K<-R(G). Therefore a<-R(G). Now recall, that the product of finite number of solvable groups is solvable. Please refer to the attachment for complete solution. Intersections of normal subgroups are exemplified.

  • Direct product

    30330 Direct Product Groups Let G1 and G2 be groups, and let G be a direct product of G1 x G2. Let H = {(x1, x2) element G1 x G2 | x2 = e} and let K = {(x1, x2) element G1 x G2| x1 = e}. (a) Show that H and K are subgroups of G.

View More Related Solutions