 In this short video, I want to introduce without proof the fundamental theorem of finite abelian groups. So let G be a finite abelian group. Then what this theorem tells us is that G as an abelian group, right? Where commutivity is required here. This group will be isomorphic to a direct product of cyclic groups of the form Z to the PIEI order. That is the cyclic groups in the factorization will be cyclic groups of prime order. Now the reason we're omitting the proof is not because it's too hard. It's just kind of a lengthy. And so if a viewer wants to see a detailed proof of this, I would recommend looking in chapter 13 of Tom Judson's abstract algebra textbook, particularly section 13.1. You can find the proof of the fundamental theorem there. So let's talk a little bit about what it means. What if our group, its order had only one prime device? You're like, let's say, for example, the order of the group was 16. What does this theorem say? Well, basically, if 16 here, since it's two to the fourth, you have to decide how are the factors of two gonna be glued together? What do I mean by that? Well, you can think of various factorizations of 16. You're using these powers of two. You could factor 16, of course, as 16. You could factor it as eight times two. You could factor it as four times four. You could factor it as four times two times two. You could factor it as, well, two times two times two times two. Let me write it that way. In which case, we get these five factorizations right here. Associate to each of these factorizations of 16. And now, these are going to be factorizations with only involving powers of two, of course. Associate to each of these factorizations of 16, we then are going to get the factors, we actually get the factorization of the group. So for example, the first one right here, we could have the group Z16, right? That's a cyclic group of order 16. We could have also the group Z8 cross Z2. We could have the group Z4 cross Z4. We could have the group Z4 cross Z2 cross Z2. And then finally, we could have the group Z2 cross Z2, cross Z2, oh boy, I need another one, cross Z2. Now, as these things can go a little bit long, if there's a repeated factor, we might write this shorter as Z2 to the fourth, right? I like this one right here, it could be Z4 times Z2 squared, something like that. But it comes down to the number of two groups of order 16, the two abelian groups, would be the different ways you can partition, you can partition 16 in this factorization, right? How many ways can you factor 16? Well, you get the five-positive factorizations. So that's what it would do if you have a group whose order is a power of a prime. Well, what do you have at different factors? Well, different prime factors, like take the group, take the abelian groups of 540. Well, 540 factors as two squared times three cubed times five. And so if you wanna think of how many ways can you get a group, an abelian group of order four? Well, two squared is equal to four or it's equal to two times two. And so there's gonna be two abelian groups of order four. You get either Z4 or you're gonna get Z2 cross Z2 with the climb four group. On the other hand, if we think of just groups of order, in this case, 27, three cubed, what can we get? Well, 27, three cubed, it's 27. You could also factor it as nine times three. You could also factor it as three times three times three. And so this gives us three possible groups, abelian groups of order 27. You're gonna get Z27. You're gonna get Z9 cross Z3. And you're gonna get Z3 cross Z3 cross Z3 or in other words, Z3 cubed. So you're gonna get these three abelian groups of order 27. How many groups of order five are you gonna get? Well, there's just the one. Z5 is the only way you can do that since it's a prime number. And so when you put this together, notice that you have two, four groups of four abelian groups. You're gonna have three abelian groups of order 27 and you're gonna get one abelian group of order five. And so as you build groups, abelian groups of order 540 by the fundamental theorem of abelian groups, what it tells you is you're gonna have to grab your favorite abelian group of order four, your favorite abelian group of order 27 and your favorite abelian group of order five. And you're gonna put those together and those choices are independent of each other. So the total number of groups, excuse me, of order 540 is gonna equal two times three times one, which is six. That's where we get these numbers for. You have all these possibilities. And so let's look at them really quick. So one option is you take Z2 squared times Z3 cubed times Z5. You could switch up the two group and you get Z4 times Z3 cubed times Z5. That's the second one. Then you could switch up the three group. You could take the climb four group times Z9 by Z3 times Z5. You could instead have Z4 times Z9 crosses Z3 times Z5. Or you could switch the three group again. You could have the climb four group times Z27 times Z5. And you could have Z4 times Z27 times Z5. And so uptie some morphism. Every abelian group of order 540 will be one of these six. Now you might be wondering, is this all of them really? I mean, what about like the obvious choice, right? What about the abelian group of order 540? Or the, excuse me, the cyclic group of order 540. Is it in this list somewhere? It is, it's actually sitting right here, right? So again, this is a list up to isomorphism. That there could be different representations of these different groups, but up to isomorphism you can find it in this list. Another example, what if we take the group say Z6 times Z6 times Z15, right? You can convince yourself that that is an abelian group of order 540. It's also in this list. It's actually right here. It's why I wrote it on the left right there. So it can be a little bit challenging to try to figure out how these things are isomorphic. But the idea you really can, the idea here is you can break up this group into prime factors, right? So looking at Z15, like we've talked about previously, Z15 can break up as Z3 and Z5. Z6 can factor as Z2 and Z3. And we can do the same thing again, Z2 times Z3. And then count what are called the elementary factors. You have Z2, Z2, you have two of those. You have Z3, Z3, Z3, Z3 of those, Z5. That's how I get this right here, this elementary factorization of the abelian group. If we did the same thing, it's kind of messy, let me clean this up real quick. If we did the same thing for Z540, well, again, if we break it up into these prime factors, this is gonna look like Z4 cross Z27 cross Z5. And so each abelian group can be given into these elementary divisors that the fundamental theorem guarantees exist. And so up, and this is like, this is trying to say that abelian groups have like this prime factorization, just like integers, we can factor them into these sort of unique primes in a manner of speaking. And so we get the following. And so, again, no proof, no proof is going on here, but it's important for an Algerist student to be able to understand and classify abelian groups. We have a very nice classification theorem, which again, if you're dying to see the proof, I would say check out, check out this reference. The explanation Judson is quite good.