Re: Inverse element proving

Quote:

Originally Posted by

**BscStud** Given a, b, c elements of the set G, and G being a group relating to the * operator.

(a * b) is the inverse element of itself.

How can I prove that (b * a) is inverse to itself as well?

If you were given only a and b, then there is a way to prove this, but since you are given c as well, I am not sure. (Smile)

Given: abab = e where e is the unit.

b(abab) = b

b(abab)b^(-1) = bb^(-1)

(baba)(bb^(-1)) = e

baba = e

P.S. Questions about groups should be posted to the University Algebra forum.

Re: Inverse element proving

Quote:

Originally Posted by

**emakarov** If you were given only a and b, then there is a way to prove this, but since you are given c as well, I am not sure. (Smile)

Given: abab = e where e is the unit.

b(abab) = b

b(abab)b^(-1) = bb^(-1)

(baba)(bb^(-1)) = e

baba = e

P.S. Questions about groups should be posted to the University Algebra forum.

Thanks for your response, I will make sure to post it next time in the correct forum. Wasn't aware.

As for your response - I'm not sure how you reached the (baba)(bb^(-1)) = e ?

As if you say (baba)(bb^(-1)) = e, you need to have somehow to prove baba = e in the first place since bb^(1) is the unit, or am I missing something?

Re: Inverse element proving

Quote:

Originally Posted by

**BscStud** I'm not sure how you reached the (baba)(bb^(-1)) = e ?

This particular line I am getting from the previous line "b(abab)b^(-1) = bb^(-1)" by replacing bb^(-1) with e in the right-hand side and by using associativity to rewrite b(abab)b^(-1) as (baba)(bb^(-1)) in the left-hand side.

Re: Inverse element proving

Quote:

Originally Posted by

**emakarov** This particular line I am getting from the previous line "b(abab)b^(-1) = bb^(-1)" by replacing bb^(-1) with e in the right-hand side and by using associativity to rewrite b(abab)b^(-1) as (baba)(bb^(-1)) in the left-hand side.

Got it, but isn't making (ab)(ab) to (ba)(ba) relying on the group being commutative (it hasn't been given to be commutative)?

Thanks again.

Re: Inverse element proving

Quote:

Originally Posted by

**BscStud** As for your response - I'm not sure how you reached the (baba)(bb^(-1)) = e ?

As if you say (baba)(bb^(-1)) = e, you need to have somehow to prove baba = e in the first place since bb^(1) is the unit, or am I missing something?

Here is a different way:

$\displaystyle \begin{align*} (ab)(ab) &= e \\ ba&=a^{-1} b^{-1}\\&= (ba)^{-1}\end{align*}$

Re: Inverse element proving

Quote:

Originally Posted by

**BscStud** Got it, but isn't making (ab)(ab) to (ba)(ba) relying on the group being commutative (it hasn't been given to be commutative)?

I am not replacing (ab)(ab) by (ba)(ba); I am replacing b(abab)b^(-1) by (baba)(bb^(-1)). Note that the order of group elements is not changed: it's bababb^(-1) in both cases. Therefore, commutativity is not used, only associativity.

Re: Inverse element proving

Think I got it, thanks Plato and emakarov! :)