how can you make bx = xb without commutative law?

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

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- Apr 6th 2010, 01:58 PMrainyicemake bx = xb
how can you make bx = xb without commutative law?

(you make apply b^(-1) since b = b^(-1) - Apr 6th 2010, 02:27 PMDrexel28
- Apr 6th 2010, 05:32 PMrainyice
- Apr 6th 2010, 05:58 PMDrexel28
- Apr 6th 2010, 07:11 PMrainyice
I am learning about Group Theory now and is in subgroup session. However, the problem does not say x = b^(-1)

- Apr 7th 2010, 03:47 AMHallsofIvy
- Apr 7th 2010, 03:55 AMrainyice
- Apr 7th 2010, 09:33 AMspoon737
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 $\displaystyle x \in G$ and consider $\displaystyle xbx^{-1}$. What is the order of this element? What does that tell you about its relation to the element b?