 Now before we carry on and do something a bit more serious I suppose I love these puzzles and this is to a few more of them and in this instance Let's just concentrate on a couple of inverses They'll come in you'll have a look so a few more puzzles just to have fun Just to get ourselves used to how to deal with these elements of a group So I have given that a b equals e so my operation a for some a b elements of this group and this group under this group under multiplication I want to show that if a b equals e that implies b a equals e so how do we prove that well we say that's given given fact that a b equals e now let's just do this we'll say b let's just do b a b and so the b goes in the front the b goes in the front b times the identity element is just b and let's just say b a b and then do a b inverse that is going to be b and b inverse and b b inverse is just e so on this side I have b a and b and b inverse is just e b and b inverse is e b and b inverse is e so I have if I have those two elements the group operation on those two elements of that gives me the identity element then if they commute then I would also get the identity element okay let's have a look at this little one I have now the group operation a on a b and c that equals the identity element for some a b and c elements of my group show that c a b equals e well that's quite easy given that a b c equals e I suppose then that we could say that c a b c equals c with a c there and then put the c inverse at the end and I have c a b and c c inverse which is this e equals c c c inverse which is this e and I have the fact that c a b is then the inverse element so that one is quite simple I suppose I could also just do do the following one then let's prove a b c c a b let's have a look b c a show that b c a b c a equals e well given that given that a b c equals e I can have that a b c a equals a and then I can have a inverse a b c a equals a inverse a and I have the fact that b c a b c a equals this e well I should just put q e d so another simple one okay how about that one show that if x a y equals a inverse that y a x is also a inverse for some a x and y all elements of this group of mine under this operation so how shall we go about proving that given let's just say that given x a y equals a inverse I suppose that means a y equals x inverse a inverse x inverse in the front x inverse in the front and that a is just going to be a y inverse at the end so that's x inverse a inverse y inverse right at the end there so I am that far but I need to to have a look at this I suppose that we can just say that a equals x inverse and I'll group those two with associativity and I'll use the fact that I have y a inverse I could also do then the same with these two that'll be equal y a x y a x inverse y a together with x you remember our properties some of our properties of these but now have a look at this remember that the inverse of the inverse we're showing that that is just the same so if I take the inverse of this it's the inverse of that in other words a inverse is just y a x and that's exactly what we wanted to show the QED another fun puzzle solved okay here we go we have quite a few givens here is a inverse b is b inverse c inverse for all for these a b's and c's element of my group under the set operation and I also let a b equals c and I've got to show that b c equals a and c a equals b both of those so given that given that a b what is it a b equals c let's see we're going to have that c a b c a b equals put a c in the front that's c c but that's also c inverse c which is also e so I have the fact that c a b equals e and what do I need I need a well let's do this one I need a c a so I'm going to say c a b b inverse that equals a b inverse at the end here that's just an e and b inverse is equal to b so c a equals b I've shown that and suppose these are not these are proofs given that so I can put QED and then we also have the other one that we have to show a b equals a b equals c so what did we do there so we put a c in the front let's put we need a b c so let's say a b c equals c c so I put a c at the end but that's also c c inverse given that little fact and that equals e and I need b what do I need b c so I can put an a inverse in the front a b c equals a inverse e nothing there this goes away b c equals a inverse but a inverse is also a so b c equals a these are so much fun to do if I say fun one more time but they really are okay let's show this one let's make it our last one these are fun to do you construct your own c play with mc if you get different solutions than what I or different ways to go about it than what I do if a b c equals its own inverse for these elements a b c elements of this group show that bca is its inverse so how might we go about that so given a b c equals a b c inverse I remember with all of these I mean there is this group operation by this I'm saying a b c for this a b c a b c elements of this group under this operation and if we put nothing there we just assume that this operation for our is multiplication under this instance that works best but you can set it up for addition as well anyway so we need to show that so what can we do let's put an a inverse what shall we put an a inverse there and an a and a b c a that equals I have what do we have we have an a inverse in the front and we have a b a b c it's inverse and we have an a there so that would equal these two would equal a b c a I suppose inverse and then a that doesn't help let's see if we can think of something else it's the c so this would be c inverse b inverse a inverse by our laws and I put an a inverse in the front and I put an a in the back that's going to work because now I have a inverse c inverse b inverse that's an e and I can put those two together so this is going to be c a inverse b inverse and I can put those together with b c a inverse and there we have a b c oops those two those two goes as b c a equals its own inverse QED almost didn't get that but quite easy to do at the end