Proving a Subgroup: Homework Statement

[STEMucator]

Homework Statement

I got this question from contemporary abstract algebra :

http://gyazo.com/7a9e3f0603d1c0dcfde256e7b05276cd

Homework Equations

One step subgroup test :
1. Find my defining property.
2. Show that my potential subgroup is non-empty.
3. Assume that we have some a and b in our potential subgroup.
4. Prove that ab^-1 is in our potential subgroup.

The Attempt at a Solution

1. Defining property : xh = hx for x in G and for all h in H.
2. C(H) ≠ ∅ because the identity element e is in C(H) and satisfies xe = ex.
3. Suppose a and b are in C(H), then xa = ax and xb = bx.

4. Show that ab^-1 is in H whenever a and b are in H. So we want : xab^-1 = ab^-1x

Start with :

xa = ax
x(ab^-1) = (ax)b^-1
x(ab^-1) = (xa)b^-1
x(ab^-1) = x(ab^-1)
x(ab^-1) = (ab^-1)x

I know this is probably horribly wrong, but for some reason I can't seem to see how to do this properly. Any help would be appreciated.

 

[Dick]

Homework Statement

I got this question from contemporary abstract algebra :

http://gyazo.com/7a9e3f0603d1c0dcfde256e7b05276cd

Homework Equations

One step subgroup test :
1. Find my defining property.
2. Show that my potential subgroup is non-empty.
3. Assume that we have some a and b in our potential subgroup.
4. Prove that ab^-1 is in our potential subgroup.

The Attempt at a Solution

1. Defining property : xh = hx for x in G and for all h in H.
2. C(H) ≠ ∅ because the identity element e is in C(H) and satisfies xe = ex.
3. Suppose a and b are in C(H), then xa = ax and xb = bx.

4. Show that ab^-1 is in H whenever a and b are in H. So we want : xab^-1 = ab^-1x

Start with :

xa = ax
x(ab^-1) = (ax)b^-1
x(ab^-1) = (xa)b^-1
x(ab^-1) = x(ab^-1)
x(ab^-1) = (ab^-1)x

I know this is probably horribly wrong, but for some reason I can't seem to see how to do this properly. Any help would be appreciated.

You skipped a step. You need to show that if bx=xb then b^(-1)x=xb^(-1). You can't just assume it.

 

[STEMucator]

You skipped a step. You need to show that if bx=xb then b^(-1)x=xb^(-1). You can't just assume it.

Ahhhh so it would be a sort of two step thing here.

First we want to show xb^-1 = b^-1x. So :

xb = bx
b^-1xbb^-1 = b^-1bxb^-1
b^-1xe = exb^-1
b^-1x = xb^-1

Now we can show that xab^-1 = ab^-1x. So :

xa = ax
x(ab^-1) = a(xb^-1)
x(ab^-1) = (ab^-1)x

 

[Dick]

Ahhhh so it would be a sort of two step thing here.

First we want to show xb^-1 = b^-1x. So :

xb = bx
b^-1xbb^-1 = b^-1bxb^-1
b^-1xe = exb^-1
b^-1x = xb^-1

Now we can show that xab^-1 = ab^-1x. So :

xa = ax
x(ab^-1) = a(xb^-1)
x(ab^-1) = (ab^-1)x

Yes, that's it.

 

[STEMucator]

Yes, that's it.

So we could conclude that C(H) ≤ G as desired. Thanks a bundle for your help :)