Register to reply

Direct product

by barbiemathgurl
Tags: product
Share this thread:
barbiemathgurl
#1
Jul29-07, 06:47 PM
P: 12
Let k be a positive integer.

define G_k = {x| 1<= x <= k with gcd(x,k)=1}

prove that:
a)G_k is a group under multiplication modulos k (i can do that).

b)G_nm = G_n x G_m be defining an isomorphism.
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
matt grime
#2
Jul29-07, 07:22 PM
Sci Advisor
HW Helper
P: 9,396
What have you done for b)? There is only one possible way you can think of to write out a map from G_nm to G_n x G_m, so prove it is an isomorphism. Remember, G_n x G_m looks like pars (x,y)....
Kummer
#3
Jul29-07, 08:22 PM
P: 291
We can use the Chinese Remainder Theorem on this one.

Define the mapping,
[tex]\phi: G_{nm}\mapsto G_n\times G_m[/tex]
As,
[tex]\phi(x) = (x\bmod{n} , x\bmod{m})[/tex]

1)The homomorphism part is trivial.
2)The bijection part is covered by Chinese Remainder Theorem.

mathwonk
#4
Jul30-07, 12:50 AM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,486
Direct product

but the point is to prove that theorem.


Register to reply

Related Discussions
Subgroup of direct product Calculus & Beyond Homework 1
Subgroup of a Direct Product Calculus & Beyond Homework 5
Semi Direct Product Calculus & Beyond Homework 1
Direct Product General Math 13
Semi-direct product Linear & Abstract Algebra 14