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### Groups and Subgroups

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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.

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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.

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

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