Help!

[cuti3panda]

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!

[Dave]

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).

[cuti3panda]

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

[haggy]

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

[/quote']

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.

[cuti3panda]

thanks a lot haggy!