# Thread: make bx = xb

1. ## make bx = xb

how can you make bx = xb without commutative law?

(you make apply b^(-1) since b = b^(-1)

2. Originally Posted by rainyice
how can you make bx = xb without commutative law?

(you make apply b^(-1) since b = b^(-1)
This question is not well-formed at all. If $x,b$ are arbitrary this is the commutative law.

3. Originally Posted by Drexel28
This question is not well-formed at all. If $x,b$ are arbitrary this is the commutative law.
exactly ..... that's why i could not solve it ...

4. Originally Posted by rainyice
exactly ..... that's why i could not solve it ...
Is this supposed to be a group? Then, $x=b^{-1}$

5. I am learning about Group Theory now and is in subgroup session. However, the problem does not say x = b^(-1)

6. Originally Posted by rainyice
how can you make bx = xb without commutative law?

(you make apply b^(-1) since b = b^(-1)
If your group it not commutative and you are not given additional information about b and x, you can't "make" bx and xb equal. In general they aren't.

Now, what, exactly, was the full statement of the problem?

7. Originally Posted by HallsofIvy
If your group it not commutative and you are not given additional information about b and x, you can't "make" bx and xb equal. In general they aren't.

Now, what, exactly, was the full statement of the problem?
the center Z(G) is defined as Z(G) = {a E G | ax = xa for every x E G}
Prove that if b is the only element of order 2 in G, then b E Z(G)

8. Originally Posted by rainyice
the center Z(G) is defined as Z(G) = {a E G | ax = xa for every x E G}
Prove that if b is the only element of order 2 in G, then b E Z(G)
This is a WAY different question than what you originally asked. It's asking you to prove that if b is the only element of order 2, then it commutes with every element of G.

Let $x \in G$ and consider $xbx^{-1}$. What is the order of this element? What does that tell you about its relation to the element b?