In a group, prove that (ab)^-1=b^-1a^-1. Find an example thats hows that it is possible to have (ab)-2=/=b^-2a^-2 Find distinct monidentity element a and b from a non-Abelian group with the property that (ab)-1=a^-1b^-1. Draw an analogy between the statement (ab)^-1=b^-1a^-1 and the act of putting on and taking off your sock and shoes (shock and shoes property).

 

anyone help pls!

In a group, prove that (ab)^-1=b^-1a^-1.

 

I'd multiply both sides by (ab) and show that both sides are then equal to the identity element (since all elements in a group must have inverses).

Hi Dave, i've been done exactly wat you tried to show me above, but my professor mark it wrong, he took my 10 pts out, that's so sad

 

Here is wat i got:

(ab)^-1=a^-1b^-1 for every a,b are in G

then b^-1a^-1=a^-1b^-1 for every a,b are in G

Mutiplying both side by abon the left: e=aba^-1b^-1

Mutilplying by right: ba=ab since this is true for every a,b are in G, G is Abelian, if G are in Abelien, then (ab)^-1=a^-1b^-1 fore very a,b are in G.

 

give me some hints Dave, thanks a lot

(ab)^-1=a^-1b^-1 for every a' date='b are in G

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That's not the case unless G is Abelian

 

c^-1 is unique in G so all you need is to show that (b^-1 * a^-1)ab = e = ab(b^-1 * a^-1), which is simple.

thanks a lot haggy!