# Inverse element proving

• Apr 13th 2013, 10:33 AM
BscStud
Inverse element proving
Hello there,

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?

Tried a lot of methods but all of them seem very far fetched (with all ending at (a * b) = (b * a)).

Any help would be much appreciated.
• Apr 13th 2013, 11:13 AM
emakarov
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.
• Apr 13th 2013, 11:23 AM
BscStud
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?
• Apr 13th 2013, 11:30 AM
emakarov
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.
• Apr 13th 2013, 11:34 AM
BscStud
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.
• Apr 13th 2013, 11:38 AM
Plato
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:

\begin{align*} (ab)(ab) &= e \\ ba&=a^{-1} b^{-1}\\&= (ba)^{-1}\end{align*}
• Apr 13th 2013, 11:40 AM
emakarov
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.
• Apr 13th 2013, 11:43 AM
BscStud
Re: Inverse element proving
Think I got it, thanks Plato and emakarov! :)