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Groups and Subgroups


Date: 10/20/2001 at 21:55:16
From: Carolyn Thall
Subject: Abstract algebra - groups

Let G be a group and let a be one fixed element of G.  
Show that H = {xEG|xa=ax} is a subgroup of G.  
The E= belongs to and the H should have a sub a. I didn't know how to 
type those on my keyboard. I don't know how to start this problem.

Another question I'm having a problem with is: 

For sets H and K we define the intersection H/\K by
    H/\K = {x|xEH and xEK}
Show that if H is a subgroup of G and K is a subgroup of G, then H 
intersect K is a subgroup of G.

Please help me - I'm lost!


Date: 10/21/2001 at 00:08:30
From: Doctor Paul
Subject: Re: Abstract algebra - groups

To show that a subset H of a group G is a subgroup, you need to show 
three things:

1. that the identity element of G is in H.

2. that if x is in H then x^-1 is in H.

3. that if x and y are in H then x*y is in H.

There is a shorter way to show that H is a subgroup. Sometimes this 
shorter way is referred to as the "One step subgroup test."  The 
reason for the name is because you only have to show one thing and it 
is equivalent to the three conditions above.

1.  If x and y are in H, then x*y^-1 is in H.


>Let G be a group and let a be one fixed element of G.  
>Show that H={xEG|xa=ax}is a subgroup of G.  

Here, H is the set of elements in G that commute with a. This subset 
actually occurs so frequently that it has a special name.  It is 
called the Centralizer of a in G (where a is a fixed element of the 
group G). This is generally denoted C(a).

To show that H = C(a) is a subgroup of G, we first show that the 
identity element of G (denoted by e) is in H. Since a is a fixed 
element of G, it is trivial to show that e*a = a*e. This follows from 
the definition of an identity element.

Now suppose x is in the centralizer of a in G. Then x*a = a*x (I'm 
just applying the definition of what it means to be in the 
centralizer)

Then:

x*a = a*x

a = x^-1 * a * x

a * x^-1 = x^-1 * a

x^-1 * a = a * x^-1

and this shows that x^-1 is in C(a).

To complete the proof, suppose x and y are in C(a). Then (again by the 
definition of what it means to be in C(a)) we have:

x*a = a*x    and     y*a = a*y

We want to show that x*y is in C(a).  That is, we want to be able to 
establish the following:

(x*y)*a = a*(x*y)

But this is exactly the case:

(x*y)*a

x*(y*a)

x*(a*y)

(x*a)*y

(a*x)*y

a*(x*y)

Thus x*y is an element of C(a) and we're done with the proof that C(a) 
is a subgroup of G. Notice that this proof is independent of the 
choice of a.

C(a) will not generally be the same set for different values of a.  
But it will always be a subgroup of G.

>For sets H and K we define the intersection H/\K by
>    H/\K={x|xEH and xEK}
>Show that if H is a subgroup of G and K is a subgroup of G then H 
>intersect K is a subgroup of G.

If H is a subgroup, it must contain the identity element of G. 
Similarly, K contains the identity element of G. Thus H interset K 
contains the identity element of G.

Now suppose that x is an element of H intersect K. Then x is in H and 
x is in K. Because H is a subgroup, x^-1 is in H. Because K is a 
subgroup, x^-1 is in K.  Thus x^-1 is in H intersect K.

We're almost done. Do you see that this problem is really quite easy 
once you approach it the right way?

Finally, suppose x and y are in H intersect K. Then x is in H and x is 
in K and y is in H and y is in K.

Since H is a subgroup, x in H and y in H implies x*y is in H. 
Similarly, x in K and y in K implies x*y is in K. Thus x*y is in H 
intersect K and we're done.

A final note. Proofs like these are not difficult. You are only given 
enough information to do one thing. Learning to do this one thing will 
go a long way in advanced math courses. I'm referring to following the 
given definitions. In these problems, that's all you have to work 
with. 

You just need to understand the properties that distinguish the 
objects with which you are working. In the first example, this 
property was that elements commuted with a given element. In the 
second example, the property was that if an element was in H intersect 
K then it was in H and it was in K. Of course we also used the fact 
that H and K were subgroups, but we didn't have any other information 
to work with and frankly, there isn't any other way to approach 
problems like these.

If you'd like to talk about this some more, please write back.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Logic
College Modern Algebra

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