cuti3panda Posted September 25, 2004 Share Posted September 25, 2004 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! Link to comment Share on other sites More sharing options...
Dave Posted September 25, 2004 Share Posted September 25, 2004 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). Link to comment Share on other sites More sharing options...
cuti3panda Posted September 25, 2004 Author Share Posted September 25, 2004 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 Link to comment Share on other sites More sharing options...
haggy Posted September 26, 2004 Share Posted September 26, 2004 (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. Link to comment Share on other sites More sharing options...
cuti3panda Posted September 26, 2004 Author Share Posted September 26, 2004 thanks a lot haggy! Link to comment Share on other sites More sharing options...
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