Group Theory II - Simple Groups Part 1

[vc_row][vc_column][vc_video link="https://www.youtube.com/watch?v=bcWnGvR7eDk"][/vc_column][/vc_row][vc_row][vc_column][vc_column_text][/vc_column_text][/vc_column][/vc_row][vc_row][vc_column][vc_column_text]okay do you still remember what a simple group is is it okay what are simple groups a group is simple if it has no no non-trivial no non trivial normal subgroup okay let g be uh d is simple if d has no none three will if you don't remember what this means okay now we have a group g right oh proper this is important proper non-trivial normal circle okay so this is a group g if you have a sub row and you have just a subgroup containing the identity right this is called the trivial subgroup this is called in proper if this is not equal if this is not equal this is proper none trivial okay because you can have proper trivial if g is not e if g is not e meaning e plus plus okay this will be proper but trivial right d is also the sub group of g because g is called the improper subgroup of itself we want the proper okay so that's why i said proper non-trivial cannot be just e normal what does it mean by normal so we have h is normal energy if a h is h a for all the less cosets is equal to the right coset all left process are equal to the right coset so h is number one so this is a simple group right so you're given a group you see whether it has a normal subgroup or not and whether the normal subgroup is proper non-trivial once you find one it's not simple okay if the only normal subgroup is this or this then it is simple okay in other words g is simple if it's normal sub groups are just the trivial subgroup and it's of the improper right so the key is to check whether a group is simple or not try to find the normal sub group okay if there is a normal subgroup which is proper non-trivial not simple okay if you cannot find all these subgroups are not normal so the only normal will be just the trivial and itself then it's simple so the key is you have to find the normal next one we see some examples of theorems i'll call it a theorem or if you don't think that it's too big property you can call it proposition okay i call it theorem all obedient groups are not so you have to get used to proving in general okay you cannot give a proof in general given an example or specific okay it won't cover all okay now how do you prove this let g be in obedient then billion group means what gh is kgg for on hg g right um yeah thus gh right is hgg for all this is the definition of normal right so if h is a subgroup of g and h is okay so this is the proof actually that all subgroups of an ability group are normal this is actually in undergraduate in modern algebra right so if g is a billion every element commutes the h is hg hc and g so if you have any sub group h gh is also hg because the definition of gh is like this right okay so you can just do this so if h is a sub group then h is normal so what does it mean it means all the billion groups are not simple because all of their sub roots are normal that's the key if you have a subgroup that is normal not simple okay so that's what i put there if all subgroups of any group are actually normal so this is the proof what is it normal okay so it's not simple the second one is if you have a subgroup a proper subgroup h of g with index 2 then g is not simple if z is a group h is the circle of g such that this is okay if this is a group h is a sub row of g with index two then z is not symbol okay this is from previous course okay if if the index is two you only have two cosets right what is h one is h you only have two coset the index is two the index wins the order of g over the order of h and by lagrange theorem is actually the number of cosets okay so you only have two cosets h is h the right coset h is h the other one what's left is h a so h has to be equal to h a h has to be equal to h a so h must be normal in g so if h is normal in g thus g is not simple once you have a normal subgroup automatically not simple okay okay okay okay next theorem three all cyclic groups of prime orders are simple didn't say all sixty groups but it says all secret groups of rhyme why you have to find a normal sub group right okay you remember we said if g is a group of prime order then g is a billion right if g is a group of prime order then g is a billion that is one way since that g so doesn't mean cyclic groups it can be any any group right any group of prime order but i just put cyclic here yeah say let's see i just put the secret room off right okay i put this actually you can put generator right if you want okay let g be a secret group of prime order then if h is a subgroup of g here it is by lagrange theorem okay what does lagrange theorem say can somebody um teacher uh order of each most divided order of j right so because because order of j um prime so not uh not any divided only j or only b or one so not any sub group of uh j or derby and brian for brian okay so that's right by lagrange theorem the order of h divides the order of g now g is a group of prime order so the order is p so what are numbers that divides p a prime number only 1 or right that's the only number that divides the prime p so it's only one or p if the order is p is the group is the group g if the order is one is the three so the order of h is one for p thus h is this or h is g okay so this means that g is simple okay so it's just by looking at the subgroup is it normal or not if there is no proper number then it is not simple if it is if if there is no that is simple sorry if there is if there is proper normal it's not simple okay that's the key okay now we go to okay theorem 1.4 is exercise exercise means make sure you try this and i can ask in the next class okay let me write error and a billion growth is simple if and only if it is finite end of prime point okay okay so i will not write the proof here but i will just give some hints here so you will put let g be a group okay then you show this one if g is a billion if z is a billion and simple right then g is higher and s okay you assume g is billion and simple meaning there is no i will give you the answer there is no normal sub rule right so you have to show this you can show by contradiction right you assume the order is not prime and then show that it's a normal sub-group meaning it's not simple a contradiction okay and you can assume this way let g be finite end of prime order then if g is a billion it is simple okay so you go the other way around okay you try this okay now next is zero five zero five six 2 on this proof that's another one that you need to find okay i didn't give the proof there because i want you to find it and try to understand okay i if you can type then it's easy for you to share in the next class if you write you have to scan so if i ask okay vian can you share the proof then you can say okay and we are in the same whatsapp group so if you need someone to check your answer or compare or discuss you are very welcome okay i'm sure you don't mind right to contact each other okay just just know each other and try to discuss okay so what does it say suppose suppose g is a group of order two times n and n is odd n is bigger than one then g is simple it's not simple the easiest okay besides false they are fosia because you have i have heard your voice i want to hear someone else's voice okay can someone give me an example of the smallest non-ability group is s3 is the smallest none per video you know permutation group of other six yeah okay so you have to remember this is from undergraduate the list of elements right one one two one three two three one two three one three two mutations right permutations are one two one and on two so the order is six which is two times three so by these two or tests it is not simple but if we don't use this theorem you can actually find a normal sub row right what is the normal sub group one with index two right so we have done this before many times this is a normal subgroup of order three so the index is two the other three sub groups of other two they are all not normal right remember this is also what we call a silo silo three of s3 right that's why you need last semesters dots with you you cannot keep it anywhere else you have to remember all this i assume you you know this so by definition this is a normal subgroup which is not trivial and proper so by definition you have a trivial not trivial okay you have a non-trivial proper normal subgroup ah that's the right one so it is not normal by this but by this much much easier right the order is six which is two times three three is odd bigger than one so it is not simple okay done the one that i give in the example there is a group of other 30 which is 2 times 15 again 2 times 15 right but this one is 2 times 15 so by two or tests it's not simple if you write this 2 times 15 is 2 times 3 times 5 if you go back to last semesters when you learning about silo theorems right you have cylo2 three cello five and it turns out that there is actually a normal sub loop right in the in this one so actually using definition also okay the last thing that i want to do today is the last theorem for today so many theorems now first day already six students right okay right okay which we call index theorem okay let's see we finite group so the order is less than infinity okay h is a proper subgroup of g sometimes people put this it's not equal is proper such that the order of g does not divide the index of h in g factorial okay then h okay h is the one page contains uh none reveal normal sub group of g okay so in particular in particular g is not if you can find the proof that would be nice yeah i'm not expecting you to write it out or show it but it's quite long but it's good for you to try and find it and see if you can understand the the okay let g be a group of other 20 so this is the example yeah okay this then blah blah blah g is not simple this is actually okay now example is the order of g is 20 the order of h is five okay if you have 20 you know that is several over the five because silo right remember cello this is four times five prime power okay you have a cell number of four your percent or whatever five so definitely you have a sub group over five okay now let's see this let's see this what is this g h is 20 over 5 is 4 and g h factorial is 4 factorial which is 16 so 20 which is the order of g does not divide 16 which is right 24.nmake sure you are awake that's what the teacher say right you make a mistake i make sure you are awake are you all sleeping okay i have a very good student here all right so g is not simple all right so this is what we will see today and some more examples i will give on wednesday all right in our class actually this notes i i wrote with um what is the one in the seminar and the hazira and group theory one i wrote without in the seminar just now so we bring up together the lecture notes<br><!-- wp:image 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