Solved

Abstract Algebra: Direct Products

Posted on 2008-10-29
4
2,165 Views
Last Modified: 2012-08-13
I have two problems that i need help with! I'm trying to study for an exam and cannot figure out these review problems

first problem:
The group S3 (+) Z2 is isomorphic to one of the following groups: Z12,
Z6 (+) Z2, A4, D6. Determine which one by elimination.

I know to be isomorphic denoted phi from a group G to a group G' is a one to one mapping from G onto G' that preserves the group operation. So phi(ab)= phi(a) phi(b). I also know to prove one is isomorphic, need to have mapping (define a function phi from G to G'), need to be 1 to 1 (assume that phi(a)=phi(b) and prove that a=b), needs to be onto (prove for any element g' in G', find an element g in G such that phi(g)=g'), and prove that phi is operation preserving (show phi(ab)= phi(a)phi(b)).

I also know that S3 (+) Z2 the first group S3 is multiplicative and second group Z2 is additive so dealing with mixed binary operations.
I think S3 (+) Z2 = {(1), 0), ((1 2), 0), ((1 2 3), 1)}
then ((1 2, 0) * ((1 2 3), 1)= ((1 2)(1 2 3), 1)= ((1)(2 3 1), 1). But i'm not too sure.

second problem:
Find a subgroup of Z12 (+) Z4 (+) Z15 that is of order 9.

I know that the direct product is the set of all n-tuples for which the ith componenet is an element of Gi and the operation is componentwise.

I'm confused how to find direct products, can someone help me? Thanks so much, greatly appreciated!
0
Comment
Question by:dongyowlin
  • 2
4 Comments
 
LVL 84

Accepted Solution

by:
ozo earned 500 total points
Comment Utility
A4

Z3 (+) Z1 (+) Z3
0
 

Author Comment

by:dongyowlin
Comment Utility
how does A4 equal to Z3 (+) Z1 (+) Z3?
A i know means alternating group. but im still confused...

0
 
LVL 84

Expert Comment

by:ozo
Comment Utility
A4 is isomorphic to  S3 (+) Z2

Z3 (+) Z1 (+) Z3 is a subgroup of Z12 (+) Z4 (+) Z15 that is of order 9.
0
 
LVL 20

Expert Comment

by:thehagman
Comment Utility
Unfortunately, all candidates have the correct order (12), so this cannot be used for elimination.
Since S3 is not abelian, the direct product isn't abelian either.
This eliminates Z12 and Z2(+)Z6.
The only subgroup of order 3 in D6 consists of the rotations with multiples of 120°.
Such a rotation never commutes with any element of order 2 (i.e. reflection); this eliminates D6.

2)
Z12 and Z15 have a subgroup isomorphic to Z3 each, together they make a group of order 9.
I.e., the subgroup generated by (4,0,0) and (0,0,5) is what you are looking for.
0

Featured Post

IT, Stop Being Called Into Every Meeting

Highfive is so simple that setting up every meeting room takes just minutes and every employee will be able to start or join a call from any room with ease. Never be called into a meeting just to get it started again. This is how video conferencing should work!

Join & Write a Comment

Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! https://www.google.com/about/careers/students/guide-to-technical-development.html How to Know You are Making a Difference at EE In August, 2013, one …
Have you ever thought of installing a power system that generates solar electricity to power your house? Some may say yes, while others may tell me no. But have you noticed that people around you are now considering installing such systems in their …
Here's a very brief overview of the methods PRTG Network Monitor (https://www.paessler.com/prtg) offers for monitoring bandwidth, to help you decide which methods you´d like to investigate in more detail.  The methods are covered in more detail in o…
Polish reports in Access so they look terrific. Take yourself to another level. Equations, Back Color, Alternate Back Color. Write easy VBA Code. Tighten space to use less pages. Launch report from a menu, considering criteria only when it is filled…

772 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

11 Experts available now in Live!

Get 1:1 Help Now