# Hello i need help with modern algebra

• Sep 16th 2005, 03:19 PM
pisey guy
Hello i need help with modern algebra
Dear All

I've been busy doing hardwork on modern Algebra On "group and ring
theory".
It's seem new to me! i try to read the book "Contemporary Abstract
Algebra: by Joseph A. Gallian".

I try to solve these following questions(I think it may be simple for
people that already took this course befor).
The questions are in the following ordered:
1. Suppose O:G--->H is a homomorphism.(for the O notation it actually
Thih but i can't write in correct notation due to board).
Suppose that K<|G with K |^| Ker(O)=1 .Show that K is isomorphic to a
subgroup of H.( where <| stand for normal subgroup,,,,and |^| mean
intersect (Khmer word Prosob.)).
2.Consider a rectangle box of size a * a * b where a # b.(* stand for
multiply,,,and # stand for Not equal to).Describe the rotational
symmetry group of this box (it is isomorphic to the dihedral group),and
use Burnside's theorem to find the number of rotationally distinct ways
of colouring the edges using three colours.
3.Assume G/Z(G) is cyclic.show that G = <x> Z(G) for some x belong to G
and hence show that G would have to be Albelian.This is ofcourse a
contradiction,proving that it is not in fact possible for G/Z(G) to be
cyclic.
4.Let G be a group of order 60,and let S be the set of Sylow
3_subgroups.
Define the action of G on S by conjugacy and explain how it can be used
to generate a group homomorphism Phi:G--->Sn where n=|s|.
Explaine Why Im(Phi)#1.
(phi actually(phi notation that phi=3.1416...) Sn i mean S sub n and
ofcourse
|s| is absolute value of s,,,,# stand for not equal to).
5.Use the result from question 4 to show that a simple group of order
60 is isomorphic to A5.

Any idear Anyone?.....

Kind Reguard
Pisey Guy
• Sep 17th 2005, 10:41 AM
rgep
For 1, you want to prove that K is isomorphic to a subgroup of H. So you need to exhibit an isomorphism between them. The only map you've got to work with is Theta. Consider restricting Theta to a map from K to H. This is an isomorphiosm between K and the image im(Theta) if and only if it is one-to-one (injective). A homomorphism is injective if and only if it has kernel equal to {1}. What can you say about the kernel of the restriction of Theta to K?
• Sep 17th 2005, 11:13 AM
Rebesques
Quote:

3.Assume G/Z(G) is cyclic.show that G = <x> Z(G) for some x belong to G
and hence show that G would have to be Albelian.This is ofcourse a
contradiction,proving that it is not in fact possible for G/Z(G) to be
cyclic.
Let's see if I understood. :eek:

We want to show that G=<x>Z(G)={x^m*z: m integer, z in Z(G)}. If this is want we want, do read the following:

Since G/Z(G) is cyclic, consider a generator xZ(G) (Caution: x is not in
Z(G)). Since we have a quotient group at hand, the cutest way to establish this equality is possibly the first homomorphism theorem. We need an isomorphism φ: G --> < x >Z(G) (greek installed in keyboard, hehehe) with kernel Z(G). This appears obvious -- Since G/Z(G)= < xZ(G) >, we have for all g in G, that there exist an integer m and z,z' in Z(G), such that gz=x^m*z'. Let φ(g)= x^m*z'. It is easy to prove that φ is well defined and 1-1 (hint: use Caution!), and also that it is onto. It's kernel is ofcourse Z(G).

Let's show that it is also an homomorphism. Consider g,g' in G. Then there exist z,z',z`,z`` in Z(G) and integers m,n such that gz=x^m*z' , g'z`=x^n*z``. Then φ(gg')=x^(m+n)*z'z``=x^m*z*x^n*z``=φ(g)φ(g'), if we remember what Z(G) stands for.

And so, by the first homomorphism theorem, G/Z(G) = < x >Z(G).

Now, for g,g' in G and z,z',z`,z``,m,n as previously, we have gg'=x^(m+n)*z'z``(zz`)^{-1}=g'g, and so G is Abelian. This is not true in general, and so G/Z(G) cannot be always cyclic.

(.......how can algebraists put up with all of this?? :confused: )
• Sep 17th 2005, 09:47 PM
beepnoodle
I have that book. What chapter and problem are you looking at?