PSE Lectures - Ep 6 - Cyclic groups, Cyclic SubGroups & Finite Abelian Groups

Transcript

this call is now being recorded okay let's go all right hi guys and sorry for the delay welcome back um we are now using a Google Hangouts mainly because it has this uh raised hand feature which I think will uh will help us coordinate so you are welcome to uh still like write questions in the chat and so on but the best way is is probably to to raise a hand then I I hear a ping and I can I can address it

right so less time uh we talked about cyclic groups and uh um we have this notation so G is a group foreign you have an element a lowercase G in the group and G and you

you can you can look at all the powers of uh of this element and this is the set of all G to the n where n is some integer could be negative where I I remind you that if if n is is negative I mean G to the minus n is simply G to the minus 1 to the n this this is by definition a and this this subset this is a subset of G of

course and this is in fact a subgroup um moreover if G is finite then a the the subgroup generated by by G can be written in fact is a g to the n where n is in

the natural numbers and in fact this is going to be some finite set because G is fine right so it's going to be e which is g g to the zero up to G to the the k where um or actually due to the K minus one let's say K is what is called the order of of the element G I mean it's the the minimal natural number such that g to the power of this number

is is again the unit element okay we saw we saw an example EG you take integers modulo 2 times integers module 3 and this is generated by the element one one I mean if you add one one to itself uh I mean six times you get you get 0 0 which is the the neutral element in this uh so so this group is is uh is cyclic

um we had uh we had a proposition that says that if G is a c click of order n then G is isomorphic to the the group of integers modulo n of course the group of integers modulo n is cyclic because it is generated by by one right you you add one to itself and

times you get the zero element this is and this is if you think of it this is kind of a classification because it tells you that um some notion that is abstractly defined as cyclic at least in the finite case is always of the form integers modulo and for some n we had a proposition that says that the number of of

of generators the number of a distinct generators of the N is what is called the Euler function VN which is all the all the numbers k smaller than n such that gcd and K is one this has a uh an immediate implication

if n is some prime p then any any X in ZN which is not zero I should say zp is a generator okay and and so you you can think of zp is a um

I mean in in zp we we use additive notation so instead of G to the N I write Ng but it would be like a e G 2G up to a p minus one G and you can do it for energy okay and this uh this gives you a applications in in cryptography in fact

and when you use a the additive group of integers module or prime you can choose a basis and this is if you can call it the the basis of logarithm because the the hard problem is uh is to solve the the so-called discrete logarithm problem that is I give you I chose an element G as a generator and and now I um I give you an element X so you know that X is equal to n g for some n but

I mean to find n explicitly if p is is uh is a large Prime to find n such that X is equal to Ng is a is assumed to be a hard problem okay and so what you can do you can choose any any gso generator and this will be the the basis of your logarithm problem so it gives you flexibility Okay so maybe in this context it's worth

mentioning uh small proposition a subgroup of a cyclic group easy click so so suppose G is of order n no g c click and take take some subgroup h of G

I can write G generated by some some element lowercase G with this element G I can I can express all the elements of age in particular because I can express all the elements of G so age is is actually um the unit element which we can think of as G to the zero and then you have G to the K1

up to G to the k D minus 1. so let's say [Music] the order of H is is G s right is d what I'm doing here I do not know that age is a g to the 1 up to G to the D because that that's really not necessarily the case um but but age must be a h has the the elements so and each of them must be a

power of of the generator so I just I just named the powers K1 up to K D minus one now D is the order of H it must divide n so we can write n equals KD for some k since

the order of age is d if I take any element in in the group H and raise it to the power of d i I have to get the the neutral element so G to the k i to the D must be the unit element this is of course G to the k i d this implies in fact it's equivalent

to say that a k i d is some um multiple of n some m i right because that's that's the order that's the order of of GD I mean the order of G is n because the whole group is generated by generated by um but then I have k i d

equals n m i but n is KD so this is KD MI and now I can I can cancel out the D and this this gives me that k i is equal to k m i thus each element in age

is a power of G to the k uh right because I um I mean Ki is is k k times m to the I and k i is the is the power of the the element G I mean G to so so I can simply write age as the unit element due to the k G to the K Square and so on

and this this of course means that age is generated by G to the k any questions about this proposition we saw that that a integers module two times integers modulo 3 is cyclic but

what is the general case so the following Lemma gives you a starting point if G and H uh groups G in G is order in in aging age as order m

the order of this element G comma h is simply the least common multiplier of Eminem so this in in the Cartesian product G times h yeah

sorry can you remind me what a GPS is so you say if G and H are GPS this is my my shortened uh writing for groups so I I say group like this okay okay thank you uh yeah I will try to to uh not use a lot of a lot of these shortings but but for some keywords I yeah it's fine I was just taking notes and I got uh like I I was yeah okay so how do we prove this this is

this is really simple say the the well I take G comma h and I raise it to the nth power in G times h to to do the operation is is to Simply do it coordinate wise so to say that this is equivalent or this is equal to the the neutral element is equivalent to say that g to the n is the neutral element in in G

and H to the N is the neutral element in H but this in turn is equivalent to say the order of G divides n and the Order of G of H also divides in so n must be a common multiple and M and I remind you the order of an

element is the the minimal number uh such that if you raise the the element to this to this number you get the the neutral element so if we we see that this number needs to be a common multiple it means that the order of GH is is the least common multiple now since we're dealing with the

LCM and gcd it's useful to to know a small fact and let's call it climber um for for two natural numbers n and M if you multiply multiple the gcd of N and M and the LCM you simply get n times m M well how do you see it

um what we do here is um we take a common a decomposition into Primes I write for example m equals T1 to the K1 up to p R to the KR and N is P1 to the L1 up to PR to the LR the Ki and Li

are bigger equal to zero so if I if I allow k i and l i to be also also zero this this allows me to to do the decomposition into primes over the same set of primes because right you you could have m equals a P1 up to p n or sorry P1 up to PS and n equals a q1 up to QT and and you have some Powers here I don't know a A1 up to a s

B1 up to BT and these are not the same sets of prime so to to get into this comp this decomposition over the same set of primes I simply allow the powers to be uh to be also zero and then I can see I can take a all the list P1 up to PS Union q1 up to QT and I can express an m and n over the same set of primes but once I do it I can I can express the the gcd and the

LCM in a uniform manner that is gcd is simply P1 to the maximum of K1 L1 PR of KR LR and and the LCM is simply P1 to the minimum

K1 L1 PR to the minimum uh okay r l r and from this it follows immediately that gcd times LCM is is n times m any questions about the Dilemma

hello it should be at art okay so what can we do with this Lemma this is the following Corollary ZN times z m is cyclic if and only if gcd

n and M is equal to one and of course in that case you have ZN times the m isomorphic to a cyclic group of order n times M there's only one so it's it's Z and M understanding so we haven't even only if here right

so let's start with the direction of and M equals to one then the order of this element one comma one which we know to be LCM is simply equal I mean LCM is equal to NM

modulo gcd and M but gcd and M is one so so this this whole quotient is is NM so we have an element of order n times M which is the order of the group right is it z n times the m is is of order n times M so so uh so basically it must generate the whole thing okay so so it is you click

conversely gcd of N and M is bigger than one the order of every element in in ZN of each element

in the m divides m for any element the order of every on each element in ZN times z m divides LCM but LCM

of N and M is by the the identity we proved this is NM modulo gcd which is strictly smaller than an m because the gcd of NN is assumed to be bigger than one and and this is a contradiction so uh well sorry not not a contradiction but but uh

but then the end times z m is not is not cyclic okay so so if if we assume that the gcd is is strictly bigger than one then all the elements have order uh strictly smaller than NM and so the group cannot be cyclic any question about this proposition Corollary

all right so now maybe a famous rotary this is called the the Chinese main Theory at least in a in a simple version for natural with equals to one Maybe

suppose and then for all r and s the the system of equations x equals r modulo n and x equals s

module m as a unique solution all right anybody has an idea how to uh how to build an argument for that I I simply build an isomorphism

weather for now I I Define a function from integers modulus NM to the Cartesian product and this map takes X into X module n and x modulo m now we actually saw if you map

if you remember we saw that if you map Z and M into for example ZN by taking a x to X module m you get a a group homomorphism and the the group operation in in the Cartesian product is is done coordinate wise so this means that that f is also homomorphism because to be a homomorphism

whose uh whose range is a Cartesian product is the same as to be a homomorphism to each component of the Cartesian product yes can you can you rephrase this or or can you like go through this again because I so I cut the inclusion of using the the summer Vision that makes sense but after you after you use it I'm not sure how you map this kind for example into the into the other group and that for example well okay I mean this uh you could simply multiple multiply the

the 2L I mean if you give me eight a b I can multiply oh I see I see I see I see I see so you will hear okay you will get the same operation in both sides and you can Traverse both sides getting the same thing all the time okay fair enough thank you this application is rejected or no yes yes yes I just say I mean the the first thing I wanted to address is that it's it's a it's a group homomorphism and I'm saying it's a group of morphism because you have actually two maps here Z and M into z n and z and M into z m

and each of them separately is a homomorphism this this was an example we we proved explicitly that each of these uh these two is homorphism and to be I hope what I'm saying to be a homomorphism into the since the Cartesian product operation is defined coordinate wise to be a homomorphism into the Cartesian product is is the same as to be a homomorphism to into each of the the components so so we know then f is a homomorphism okay and and now I claim that f is

indeed a bijective as well clearly F of one is one comma one right I mean one more to do n and one module to m since we we assume that the the gcd of N and M I I We Know by by previous uh Lemma that of one comma one is n times m

so in fact this this already solves you the the problem I mean you you mark uh an element of order uh n times m into an element of order n times m so so you get everything right we get f is bijective hence

it's an isomorphism I mean znm sorry now we we are so so this is a if you want the first part of the proof is telling you something we we talked about I mean we we knew that these two groups are isomorphic but we didn't have a concrete isomorphism and now that we have a concrete isomorphism we can solve this uh the system of equations

uh I mean the note are and to be a r model n and SN to be sorry s m to be s modulo as since f is dejective um the element

RN SN in the N times z m has a unique source X in integers modulo n times m such that equal to R to our end equal to s

work it's m okay and that's that completes the proof any question about the about this uh this result sorry daily trust.com uh yes I wanted to well yes uh I thought so we defined concretely the

F so going France CNM to to c n times cm um but the other way around it's a little bit more tricky right because what what we put two down two lines down in parentheses um that's not the inverse effect right yeah actually I haven't thought about it I think it's because you completely in many cases and they will lead us to the same element

it could be so I I you might be right uh let me let me delete it or put it in a question mark uh the universe yeah the universe is I need to think about it a bit I think or at least intuitively it looks like it it should be a times B but but let's leave it for now all I'm saying is to to see that you have an isomorphism you need to have a group map that is bijective and to see that it is bijective this this yellow box

tells you everything because you are mapping I mean the the element one his order in him right in integers modulo and M and the element one comma one by by a result we prove also has order n times M because the the gcd of N and M is is uh is one so if you mark an element of order n or order K into another element of order K between two cyclic groups you have a bijective map right because

you you are mapping I mean you have two cyclic groups let's say one is generated by by G and another is generated by age so let's see is a g zero up to GK and an age is eight zero up to h k and you define a map by saying G goes to h or if somebody gave you a map that in particular satisfies that g is mapped

into h then you know and this map is is a good is a group homomorphism then you know that this map must be a must be rejected right because it will map G square into H square and so on is this clear David yes thanks so uh yeah I'm fully convinced that it's uh by objective but the thing is I was at some point I had to to try to go the other way around being a problem I remember because a bit of a pain I think can it be so could you do ever to the minus one so be the other way around yeah yeah so as

I said I need to think about it a bit and I don't want to spend too much time so maybe after the break but yeah it should be simple okay thanks what I want to do now I mean soon we're going to take a break but but let me tell you a bit about the and the next steps so this is usually referred to as as classification of finite

obedient groups and what we mean by by classification in mathematics is you you take some some kind of abstract object like finite a billion group just a finite set with the binary operation and so on and you you give a a an exhaustive list of all the possible uh forms it can take of course since I mean since we're talking about any finite a billion group

this this list needs to needs to be some kind of rule applicative to any I mean to to integers of of any size and uh we will see eventually I mean we actually saw the beginning of this classification result by a by by the the proposition they told us that a integers modulu n times m is in fact a the Cartesian product of two uh I

mean of of integers module n and introduce modulo module m in some cases uh so so the idea generally in in this classification is to decompose and finite a billion group into a Cartesian product of of simple groups simple a billion groups and a we are now actually focusing on on a billion groups so it makes sense to change the notation I write G with additive notation

I write G plus h for the Cartesian product okay this this Plus is is coming from from this this uh this convention so instead of using multiplicative we transfer into a additive notation and since we are leaving the the kind of General realm of groups I want to mention that indeed we most of the results we proved so far were anyway about a billion groups but

were we like we stated the uh sometimes more General results and this was kind of a deductive exercise to to use the multiplicative notation uh instead of the the additive ones even when we talked about the billion groups uh in fact a main results we we proved like a koshy theorem and the first isomorphism theorem and LaGrange theorem are actually true for none of billion groups but you need to formulate them

slightly more carefully and and so now that this this is a this is done we focus on a billion groups which will anyway be uh the the the main application we have in in edit curves I guess now is a good time for a break but any questions remarks before that yeah sorry um could you just um explain again on um a previous part where you mentioned that the order of an element is n times I think

it's uh yeah it's above yeah the blue the blue ink yeah so what I'm saying here in the proof we assume that uh that n and M are core Prime I mean gcd of N and M is one and we had a previous proposition that said that in this case z ZN times z m it's like if you want the simple example is what we did last time Z2 times E3 is generated by one comma so if it is generated by by one comma

one then it means that the order of one comma one has sorry I'm sorry for this uh uh okay so this this means that the the order of n of one comma one has uh is is simply n times m right he generates a group of size n times n right so other the the definition of order means I thought it means the number of elements in a group but in this case it's different so so uh you need to there is a distinction there

is order of a group is the number of documents of a group and there is order of an element in the group so which is the least integer or the minimum natural number such that this element to the power of this this number will be will become one I mean all right okay so G is the the minimal n such that g to the n okay okay yeah and of course in additive notation it's the minimal n such that NG is equal

to zero okay and so I click the the the the previous claim we had tells it tells you that the order of one comma one in z n times z m is is uh is n times m in case n and M are are Co Prime yep yep thanks so so now we have a map that Maps an element of order n times M into into another one of order n times M it must be an isomorphism okay so

if there are no questions I suggest we meet here at the I mean 10 minutes right um so the first thing uh we can do about the classifying the finite obedient group where we can generalize this uh this result we we saw before the break so we look at at integers modulu n let and be a P1 to N1

PK to NK and this I called Prime decomposition of n so the end the N i's are strictly bigger than zero right there they are you have uh some list of primes uh to some to some list of powers and you multiply all of them you get the

integers n the integer n Now by by the previous result it take let's say s is equal to P1 to the N1 and T is equal to the rest of it so P2 to the N2 up to PK to the NK then of course gcd of of s and t is one right because they they do not have any any common uh any common Prime in the the device name and so

they have they are copyright and this means that integers modulo n is then isomorphic two integers modulus but but s is P1 to the N1 times integers modulating and by applying the same argument on t I get that this is isomorphic to integers module to P1 to the N1 or in fact we are now in an a billion or

additive notation so I write this o plus integers modulo PK to the NK okay so that's a that's a nice factorization of of integers modulo n but of course we are not guaranteed that every a billion group or every finite obedient group is is integers modulo n of course if the group is not cyclic it cannot be isomorphic to introduce modulo n

um but we can do something similar a g finite it actually doesn't have to be infinite a a billion group a p a prime we denote by GP the set of all elements in G

whose order is well such that there exists is it old a of P okay so so it could be that a to the I mean this is equivalent to say that it's all the elements in G and since we use additive notation it means that the

P to the n a is equal to 0 for sum n note that this this n could be different for different different elements a claim first the GP is a subgroup so in our annotation this is going to be a subgroup of G and why is that it's very easy I mean if

A and B are in G so in GP for example a p to the N A is zero P to the m b is zero so first the the inverse of B is minus B and this means that a p to the m minus B it which we know is equal to minus P to the m b is is again zero so

so minus B belongs to GP as well a minus B if you multiply it by P to the n p to you get 0 again so all together a minus B or a plus b as well is also an element in GP okay and and GP is called primary

subgroup and PNG we have the following Lemma it's good in a billion group an element

of finite order a as a sum A1 Plus a k the AI is our element in GPI a P1 up to Primes

appearing in the decomposition of the order of a okay and these These are uh distinct all right so how do we how do we prove

we use induction on the of distinct Primes the the divides elements it

we call this number k so if K is equal to 1 we have a only one prime that divides the order of the element a that means that um a is some Prime power P to the P to the T let's say then of course by definition a is in the the P primary subgroup GP so that's okay I mean of course it can

then it is a sum of a of the form here now suppose for every element who's in order is divisible by it most

K minus 1 Primes say P1 up to p k minus one uh I suppose that that for every such element I can write and b equals B1 Plus B K minus one with a bi an element in g p i if I if I now take

the order of this element divisible by a k distinct Primes the order of a is a P1 to the N1 has to be consistent with denote P1 to the R1 up to a p k to The RK

and now I take M to be the the tail of this decomposition so uh P2 to the R2 PK to the arcade and N I take I take to be the head of this decomposition so P1 to the R1 uh so so of course it means that I can write the order of a as simply n times m

and the gcd of N and M there is a Lemma I mean a result from elementary number Theory that says that if you have two two integers who are co-prime uh like n and M here there exists you and V in z I mean they could be negative u n plus VM is equal to one

this is a actually I want it the other way mu plus NV is equal to one this is the the so-called extended uh euclidious very euclidean algorithm I mean the euclidean algorithm allows you to to calculate the gcd of

two integers and the extended euclidean algorithm a finds coefficients u and v such that mu times times NV is equal to 1. and using this this is called by by the way this is called the zoo so u and v are called the bizu coefficients okay and so if I have this this u and v

is n times m and I remind you n is P1 to the R1 P1 to the R1 times m u a well I should say first this coefficients I can multiply all this equation by a and what I get is that m u a plus n v a is equal to a this equation

in blue is an equation of integers but the equation here in in purple is an equation of elements in G right so because G is is written in additive notation it means that I can multiply integer an element of G by by an integer uh so using this purple purple identity what I have is that a P1 to the R1 times MUA which by definition is equal to n m

I mean UA is U.S UA is an element in G or if you want a is order NM so this this is equal to to zero which means that m u a is an element of G P1 right because the I mean if I multiply it by by some there is some some power of the Prime P1 such that if I multiply mua by this this power I get 0.

also I take n sorry I take M and I multiply it by n v a what I get is V m n a but again a is is of order n times M so so this is 0 is B times 0 is 0.

NVA um well the order of NVA a p p 1 to sorry P2 to the R2 and now I can use the the inductive

hypothesis because uh has at most a k minus one distinct primes in its a prime decomposition so by by by induction by inductive hypothesis I can write n v a as as uh as a sum let's call it a A2 up to AK where the AIS are elements in the

in the primary Pi group or I equals 2 up to K but then we're done remember a is equal we had this this purple a purple box a is equal to um m u a mua we saw that it's it's an element of I mean this is like

m-u-a Plus AK and you you can simply denote this this element is a we saw that it's an element of gp1 so you can simply denote it as a as A1 okay and this this finishes the proof any question about dilemma so if you think of it it's a it's quite

powerful because it tells you that you can you take any any finite obedient group I mean here in the in dilemma it's not assumed to be finite but for Simplicity uh I mean since we only care about elements of finite order we can assume G is finite and then it tells you that you can you can express any element is a sum of of and each of these uh these components of the sum are a our elements in in primary AP groups this level has an immediate Corollary

theorem and a g b and finite 2 billion group with let's say that the order of G has a primary decomposition as a P1 to the R1 PK to The RK

then I claim that g is isomorphic to the the Cartesian product or the the so-called direct sum of all its uh P primary subgroups so just before we we get into the proof of this theorem observe that each of these components of the direct sum is a

subgroup of G so we have kind of uh a funny a funny case in which a a big a big object G is isomorphic to the Cartesian product of of some of its sub-objects before I before I go to the proof any questions about the the statement or the the previous uh stuff um I might have one which is yeah can we have so many properties for this uh for

this problem sometimes in the sense of like they have to be cyclic it is cyclic but can we assume anything about that or the like the sum of the orders of the social fluids to end up being the order of G uh can we expect any extra properties from here uh that's useful or not well I I mean uh what you can say is that a let's say we postpone the proof a bit I mean the the order the order of uh GP is is is a is p to the N for some n

how do I know this I mean if if there was another Prime I mean if P Prime would divide the order of GP then by koshy by koshy theorem I would have a an an element of order P Prime but GP the definition of GP is all elements whose order is a power of P so there cannot be any element of order P Prime for for another Prime so so this means the the order of GP must be a power of P

it does but it does not need to be cyclic okay of course it could be of it could be the GP is integers modulo P to the n but that's not no that's not the only option uh for example what what is another option you could uh you could have integers module a P times integers modulo p times integers modulo P to the N minus 2.

sorry for the here we use direct sum okay and that's these two these two groups are not isomorphic this group is cyclic right the upper one let's say the blue one is a cyclic group a or if you want here in this in this group in the low the purple one you have an element

a one comma zero comma zero whose order is p right so you you need to uh okay you need to think a bit about it but it's not going to be a these these two groups are not going to be isomorphic because you will have a you will have elements of of different order in this uh uh I mean in these two groups in the the blue and the in the purple so a priori we don't know what GP might be I mean it

could be it could be we have a kind of a restricted range but still there could be differences minus two well we we we we can map but it's not necessarily going to be an isomorphism I mean remember that Z and M is isomorphic to ZN plus z m n and M are co-prime okay fair enough okay okay okay okay

so so these these two things are are not going to be arithmorphic and but but what the theorem tells you is that whatever these these primary P groups are you can always decompose g into the the direct sum of its primary pickups oh so we take an element in G

and we know that the order of this element must divide the order of the group G the Primes in the decomposition are a subset of P1 up to

PK right these are the the the primes in the decomposition of the order of G a we can thus a is equal to A1 a i in g p p i for for I equals 1 up to K and here of course it could be that for

example P1 does not appear in the decomposition of a but then we take A1 to b0 okay so I I can always do this claim and sum in Star is unique

unique in that if I could write a as a as a different sum B1 up to up to BK with the bi being elements in the the primary Pi group then each of the the AIS would would have to be equal to the to each of the bis foreign so suppose let also a is equal to B1 b k

of course with the bi um then well I take these two equations star and double star and I conclude that A1 minus B1 is equal to B2 minus A2 b k minus a k this this element A1 minus B1 is a

difference of two elements in in gp1 and gp1 is a group so this is an element in gp1 and similarly here the the direction is reversed but it doesn't matter it's still an element of gp2 and so on this guy is an element of a g uh

if if it's an element in of GPI it means there exists s i bigger equal to zero let's say bi minus a i is a well it has an order of p i to the s i if we take

M to be the tail of this decomposition so P2 to the r r two up two a p k to The RK actually instead of R I want I want s S2 and up to p k to the SK I mean what does it mean that the order of bi minus a i is is p to the i s i it means that P to the i s i times bi minus a i

is equal to zero and for this for this choice of M I get that M times bi for every I between 2 and K but by by equation a triangle I get that M times

B sorry A1 minus B one also has to be zero so the order of A1 minus B1 a writes m which I remind you is P2 to the s to the S2 up to PK to the SK but this cannot be unless but of course

this is an element of gp1 so its order has to be a prime I mean a power of P1 and P1 it does not appear in this decomposition so the only way this can happen is that a S1 right we say if a A1 minus B1 let's say is is order P1 to the S1 S1 the only way this this can uh

can divide m is is with S1 equal to zero A1 is actually equal to B1 and you do the similar similar argument to the deuce similarly a i is going to be equal to bi for all I we're we're almost done so this is the the heart of the argument we now Define

f from g into the direct sum by taking I take an element a it has a unique decomposition A1 and I map it into the elements A1 up to up to AK in all the I mean the couple A1 up to n up to AK and uniqueness

uniqueness of the sum says that um f is uh is injective now f is rejective because if you give me any Tuple if you give me

a couple A1 up to AK then I I can I can construct the element a to be the sum of all this element of all the coordinates in the Tuple and by definition of f of a will be the stopping so for every couple in the in the Target group I have a I have a source and moreover f is clearly a homomorphism

f is a homomorphism I mean if you take which you express as a sum and B which you express is is another sum then by definition you know that a plus b is is the sum of

all these elements and I can write it as A1 Plus B1 does a K plus b k since A1 and B1 are in in gp1 so is the so is the sum this is true for all K I mean for all uh false summons so this means that the the purple the purple sum here

a decomposition of of a plus b into a sum of elements in the the primary in the primary groups and we know that this decomposition is unique so so that's fine and and this means that F of a A plus b is going to be equal to a A1 Plus B1 in the first coordinate and a K plus b k in the K coordinate and this by definition is is a A1 up to a k

up to b k which is simply F of a plus F of B okay so we we got an isomorphism about this theorem okay so we we're kind of halfway into

the uh to proving the the result of a a classification or sometimes it's called the the structure theorem for a finite a billion group for now you can see that we are already able to decompose and this is the the result here in blue we are already able to decompose a into a direct sum of its uh primary P and and now the next step is to take a primary P group and to decompose it

further but this I think it's better to to postpone it for next time before we end any questions comments uh just related to what we do what you said now uh these primary groups that come from G they don't need to have a play model right meaning they can also be decomposed yeah I mean this this is what what is written here in this observation yeah yeah because when you said primarily I I went to like Prime yeah but yeah it's kind of it's primary because it's like a

a main component in the the decomposition but it's not the crack yeah thank you other questions all right so uh um let's let's end here and then see you next week