# Homework Help: Find x in Groups

1. Feb 19, 2012

### mikael27

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

Let G be a group with identity e, and suppose that a, b, c, x in G.

Determine x, given that x2a=bxc-1 and acx = xac.

2. Relevant equations

3. The attempt at a solution

I know the three axioms for group.

G1. Associativity. For all a, b, c in G, (a * b) * c = a * (b * c).
G2. Identity. a * e = a = e * a.
G3. Inverses. a * b = e = b * a.

But how i will use the axioms to find x?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 19, 2012

### morphism

Step 1: Multiply the first equation through by c on the right. What do you notice?

3. Feb 19, 2012

### Dick

Start by multiplying both sides (x^2)a=bxc^(-1) by c on the right. See where that leads you.

4. Feb 19, 2012

### mikael27

so it will be :

(x^2)*a*c=b*x(c^-1)*c

(x^2)*a*c=b*x

this is from the first equation.

what about acx = xac.

what i have to to next?

5. Feb 19, 2012

### Dick

Use acx=xac. (x^2)*a*c=x*x*a*c.

6. Feb 19, 2012

### mikael27

so
(x^2)*a*c=b*x(c^-1)*c

(x^2)*a*c=b*x

(x^2)*a*c=x*x*a*c=x*(acx)=b*x

Thus b=acx

and x=b*a^(-1)*c^(-1) ??

7. Feb 19, 2012

### Dick

b=xac, yes. The rest no. Cancel them one at a time. The group isn't given to be commutative.

8. Feb 19, 2012

### mikael27

so
(x^2)*a*c=b*x(c^-1)*c

(x^2)*a*c=b*x

(x^2)*a*c=x*x*a*c=x*(acx)=b*x

Thus b=acx

a^(-1)*b=a^(-1)*a*c*x

a^(-1)*b=e*c*x

a^(-1)*b=c*x

c^(-1)*a^(-1)*b=c^(-1)*c*x

c^(-1)*a^(-1)*b=e*x

therefore x= c^(-1)*a^(-1)*b

9. Feb 19, 2012

### Dick

Oh heck. I read your answer wrong. b=xac. not b=acx. Try that again. Order counts here. You were almost right. Sorry.

10. Feb 19, 2012

### mikael27

so
(x^2)*a*c=b*x(c^-1)*c

(x^2)*a*c=b*x

(x^2)*a*c=x*x*a*c=x*(xac)=b*x

thus b=xac

a^(-1)*b=a^(-1)*x*a*c

a^(-1)*b=x*c

c^(-1)*a^(-1)*b=c^(-1)*x*c

c^(-1)*a^(-1)*b=x

Is it correct now?

11. Feb 19, 2012

### Dick

No. Why do you think a^(-1)xa=x?? If you have b=xac cancel c on the right first.

12. Feb 19, 2012

### mikael27

so
(x^2)*a*c=b*x(c^-1)*c

(x^2)*a*c=b*x

(x^2)*a*c=x*x*a*c=x*(xac)=b*x

thus b=xac

b*c^(-1)=x*a*c*c^(-1)

b*c^(-1)=x*a*e

b*c^(-1)=x*a

b*c^(-1)*a^(-1)=x*a*a^(-1)

b*c^(-1)*a^(-1)=x*e

Hence

x=b*c^(-1)*a^(-1)

ok now?

13. Feb 19, 2012

### Dick

You forgot to use xac=acx again. But other than that, it's ok.

14. Feb 19, 2012

### mikael27

where do i have to use xac=acx ??

15. Feb 20, 2012

### Dick

On the third line. You want to cancel x on the right.

16. Feb 20, 2012

### mikael27

cancel x? i dont understand!

17. Feb 20, 2012

### Dick

How did you conclude "thus b=xac"? If you multiply one side of your equality by x^(-1) on the right, you have to multiply the other side by x^(-1) on the right. You can't switch sides. The group might not be commutative. That's why they gave you acx = xac.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook