How can you prove that ab = ba implies ba^(-1) = a^(-1)b in Fintie Groups?

[patelnjigar]

Show that whenever ab = ba, you have ba^(-1) = a^(-1)b.

I don't know how to slove problem.

pls help me..

 

[antonantal]

On the equality $$ba^{-1}=a^{-1}b$$ make a left multiplication and then a right multiplication by $$a$$ and see what you get.

 

[patelnjigar]

sorry.. I don't understand.. Please give me clear as that.. thanks.. smile..

 

[antonantal]

You don't know what a left or right multiplication is? It means that you multiply both the members of the equality, by the same quantity, on their left/right side.

For example, a left multiplication by $$a$$:

$$ba^{-1}=a^{-1}b$$ => $$a{\cdot}ba^{-1}=a{\cdot}a^{-1}b$$

Now go on and make a right multiplication on the equation we obtained.
Also consider what is the result of $$a{\cdot}a^{-1}$$ and $$a^{-1}{\cdot}a$$

 

[patelnjigar]

Show that whenever ab = ba, you have ba^(-1) = a^(-1)b.

then you said that

just answers:
For example, a left multiplication by LaTeX graphic is being generated. Reload this page in a moment.:

LaTeX graphic is being generated. Reload this page in a moment. => LaTeX graphic is being generated. Reload this page in a moment.

 

[patelnjigar]

you mean that I have to make left and right..

make left:
ba^(-1) = a^(-1)b => aba^(-1) = aa^(-1)b

make right:
ba^(-1) = a^(-1)b => ba^(-1)a = a^(-1)ba

then what??

 

[antonantal]

you mean that I have to make left and right..

make left:
ba^(-1) = a^(-1)b => aba^(-1) = aa^(-1)b

make right:
ba^(-1) = a^(-1)b => ba^(-1)a = a^(-1)ba

then what??

No. I mean make left, then, on the result of the left multiplication, make a right multiplication. Let's see what you get.

 

[patelnjigar]

i m sorry.. I lost and I don't understand what you talk about. Please help for that.

 

[berkeman]

Thread moved from General Math to Homework forums. patelnjigar, you have received some very good help here so far from antonantal. You need to understand that we DO NOT do your homework problems for you. You need to show us more of your work on this problem, and stop trying to get it solved for you. Show us your work.

 

[antonantal]

Come on, you almost did it!

You did a left multiplication by $$a$$ on

$$ba^{-1}=a^{-1}b$$ and obtained

$$a{\cdot}ba^{-1}=a{\cdot}a^{-1}b$$

Now if you do a right multiplication by $$a$$ on $$a{\cdot}ba^{-1}=a{\cdot}a^{-1}b$$ what do you obtain?

 

[patelnjigar]

ab=ba

a^(-1).ab.a^(-1) = a^(-1).ba.a^(-1)

(a^(-1).a) b^(-1) = a^(-1).b(a.a^(-1))

e.b^(-1) = a^(-1).b.e

b^(-1) = a^(-1).b

is that right?? I hope that I made it...

 

[antonantal]

I can't figure out what is it that you don't understand. Look:
I take $$ba^{-1}$$ ,I do a left multiplication by $$a$$ and get $$aba^{-1}$$ ,then I do a right multiplication by $$a$$ and get $$aba^{-1}a$$.

You try it on the other member of the equation, $$a^{-1}b$$

 

[patelnjigar]

ba^(-1) = a^(-1)b

aba^(-1) = aa^(-1)b

aba^(-1)a = aa^(-1)ba

ab = ba

i seem that it done answers.

 

[antonantal]

That's right. It wasn't that hard was it?

 

[Gokul43201]

ab=ba

a^(-1).ab.a^(-1) = a^(-1).ba.a^(-1)

(a^(-1).a) b^(-1) = a^(-1).b(a.a^(-1))

e.b^(-1) = a^(-1).b.e

b^(-1) = a^(-1).b

is that right?? I hope that I made it...

You started off correctly but made a typo (bolded) in the second step. Else you would have got the correct answer.