# Proving a subgroup

1. Oct 15, 2012

### Zondrina

1. The problem statement, all variables and given/known data

I got this question from contemporary abstract algebra :

http://gyazo.com/7a9e3f0603d1c0dcfde256e7b05276cd

2. Relevant 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.

3. 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

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.

2. Oct 15, 2012

### Dick

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

3. Oct 15, 2012

### Zondrina

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

4. Oct 15, 2012

### Dick

Yes, that's it.

5. Oct 15, 2012

### Zondrina

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