Analysis Questions

Date: 12/07/96 at 23:57:34
From: Yue Chen
Subject: Abstract Algebra

Hi,

I'm just trying to get ready for finals and I have been going over 
some of my professor's old exams and I can't seem to figure out some 
of the questions.  I was wondering if you could verify some of my 
answers and help with a few questions.  I hope you can help.

The questions that I am having difficulty with go like this:

(1) Let G= <a> be a cyclic group of order 15:
    (a) Show that G has subgroup H of order 5
    (b) List all the cosets of H in G for the subgroup H that you 
        found in part (a)

My answer for part (a) was H = <a^3 >= {a^3, a^6, a^9, a^12, a^15 = e}
For part (b), I have no idea.

(2) Describe two subgroups H and K of (Z, +) such that HUK is not a 
    subgroup of (Z, +).
     
I think it has something to do with -1 and 1.  That's about as much     
as know.

(3) If G is a subgroup of order 15 and H is a non-trivial subgroup 
    of G, what are all the possible orders for H?

My answer was 1 and 15.

(4) Let A be a finite set with at least two elements.  Is it possible 
    to define a mapping from A -> AxA that is (i) one-to-one?  
    (ii) onto?
  
The problem that I have with this question is, I can't see how this 
can be a mapping.  Is this similary to a binary operation where       
AxA -> A?  How can a single element map to 2 elements?

(5) If x, y, z, w are integers with x = yx + w, prove that 
    (x, z) = (z, w).
     
Should I use relations to prove this?      

Date: 12/11/96 at 14:18:01
From: Doctor Ceeks
Subject: Re: Abstract Algebra

Hi,

(1) When you have a subgroup H of a group G, the group is naturally
partitioned into cosets of H, which are subsets of G of the form xH 
(for left cosets...for right cosets, you use Hx, but in an abelian
group left and right cosets are the same). So to answer (b), you have 
to look at your group G and pick an element g and look at the set gH.  
Next, find an element of G not in gH, say x, and look at the set xH.  
Next, find an element of G not in gH or xH...say y, and look at the 
set yH.  Keep going until you every element of the group is accounted 
for.

In this case, the answer is H, aH, and a^2 H.

Note: gH means the set of elements of the form gh where h is in H.

(2) This doesn't have to do with -1 and 1.  Instead, the point is 
that a union of subgroups is generally not closed under the group law.  
Consider H = 2Z and K = 3Z.  Note that 2 and 3 are in H U K, but 
2+3 = 5 is not.  Therefore H U K cannot be a subgroup of Z.

(3) The order of a subgroup always divides the order of the group.
Without any additional information, it is possible for a group to
have subgroups of every order which divides G (for example, if G is
cyclic, this will be true).  In this case, 15 = 3*5, and by Sylow's 
theorem, there exist subgroups of order 3 and 5.  So in this case, 
there are subgroups of orders 1,3,5, and 15.

(4) AxA is the set of ordered pairs (x,y) where x and y are in A.  
These ordered pairs are the elements of A x A by definition, so a map 
which maps into AxA has values elements of AxA, or, in other words,
values which are ordered pairs (x,y) with x and y in A.

Example:  A = {a,b}.

Then AxA = { (a,a), (a,b), (b,a), (b,b)}

A map from A -> AxA is the same thing as assigning to a an element in 
AxA and assigning to b an element in AxA.

For example:  f(a)=(a,b), f(b)=(b,b).

Since A and AxA are finite, the questions you are being asked are 
really questions about the relative sizes of the domain and range.

Since the order of AxA is the order of A squared, and since it is 
assumed that the order of A excedes 1, we know that the order of AxA 
is greater than the order of A. So it is possible to have one to one 
maps from A to AxA, but it is impossible to have an onto map from A to 
AxA.  (This conclusion changes in the case where the order of A is 1, 
or if A has infinite order.)

(5) I assume that (a,b) denotes the greatest common divisor of 
a and b. If (x,z)=(z,w), it means that every d which divides both 
x and z must also divide w, and conversely, every d which divides 
both w and z must also divide x.

Let d divide both x and z.  Since w = x-yx = x(1-y), we see d 
divides w.

Let d divide both w and z.  From x(1-y) = w, it cannot be concluded 
that d divides x, because some factors of d may divide 1-y only.

This suggests a way of arriving at a counterexample:
let x = 3, y = 3, w = -6, z = 2.

-Doctor Ceeks,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/