# make bx = xb

• Apr 6th 2010, 02:58 PM
rainyice
make bx = xb
how can you make bx = xb without commutative law?

(you make apply b^(-1) since b = b^(-1)
• Apr 6th 2010, 03:27 PM
Drexel28
Quote:

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.
• Apr 6th 2010, 06:32 PM
rainyice
Quote:

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 ...
• Apr 6th 2010, 06:58 PM
Drexel28
Quote:

Originally Posted by rainyice
exactly ..... that's why i could not solve it ...

Is this supposed to be a group? Then, $x=b^{-1}$
• Apr 6th 2010, 08:11 PM
rainyice
I am learning about Group Theory now and is in subgroup session. However, the problem does not say x = b^(-1)
• Apr 7th 2010, 04:47 AM
HallsofIvy
Quote:

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?
• Apr 7th 2010, 04:55 AM
rainyice
Quote:

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)
• Apr 7th 2010, 10:33 AM
spoon737
Quote:

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?