Decomposition per the Fundamental Theorem of Finite Abelian Groups

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  • #1
jstrunk
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I need examples of how to do it
According to the book I am using, one can decompose a finite abelian group uniquely as a direct sum of cyclic groups with prime power orders.
Uniquely meaning that the structures in the group somehow force you to one particular decomposition for any given group.
Unfortunately, the book gives no examples of how to do it and none of the exercises give solutions.
All I want is a pointer to some examples that go through the process slowly at a beginner level.
In particular, no Sylow stuff. At this point in the book we haven't covered Sylow yet so apparently I should be able to do this without it.
I have searched all over the internet and can't find any.
Thanks for your help.
 

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  • #3
Office_Shredder
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That uses the abelian groups can be factored theorem which mentions Sylow in it, so I'm not sure it's as useful as you were hoping.
 
  • #4
fresh_42
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That uses the abelian groups can be factored theorem which mentions Sylow in it, so I'm not sure it's as useful as you were hoping.
I think this can be done with exact sequences (without mention them).
 
  • #5
Office_Shredder
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Can you just do something like this (basically the exact sequence idea)? I'm typing on my phone so some details will be skipped.

Let ##|G| = \prod_{i=1}^n p_i^{k_i}##. Let H be the subgroup of all elements whose order is divisible by just ##p_2,...,p_n##. H is a subgroup since the order of hk divides order of h * order of k, and order of h inverse equals order of h. Let K be the subgroup of elements whose order is divisible by ##p_1##. By induction (assuming the base step for now) H is a direct sum of cyclic groups, and so is K. We need to show G = HxK.

Obviously ##H\cap K## is the identity, so the map from HxK to G of (h,k) maps to hk is injective. To show it's surjective - let ##g\in G##. Then g has order dividing |G|. In particular there is some smallest k such that ##g^{p_1^k}## has order that is coprime with ##p_1##. Let ##h=g^{p_1^k}##. Then g/h has order ##p_1^k##, call this k. g=hk, so our homomorphism is surjective as well. The
Note we also get uniqueness from uniqueness of the base case - it's clear from the magnitude of G that the cyclic groups are unique up to picking how the cyclic groups of each individual prime are distributed. But hence by induction since H and K have unique decompositions we get that G has a unique decomposition.


Then we have to prove the base case where there's only one prime factor. I'm not quite sure how to do that at the moment.
 
  • #6
jstrunk
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These suggestions are not really what I am looking for.
What I need is to take several specific non-isomorphic abelian groups of the same order, and show how to get their different decompositions based on the different structures of each group.
 
  • #7
fresh_42
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These suggestions are not really what I am looking for.
What I need is to take several specific non-isomorphic abelian groups of the same order, and show how to get their different decompositions based on the different structures of each group.
Look at ##V_4## and ##\mathbb{Z}_4## and you see the difference: the elements of different order.
 
  • #8
jstrunk
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Is there anywhere online where a lot of groups are listed along with their unique decomposition according to the Fundamental Theorem of Abelian Groups?
Maybe if I had something like that I could reverse engineer it to figure out how they derived the decompositions.
I have seen some places where various groups are listed along with some groups they are isometric to, but it isn't clear which isometry is the unique decomposition required by the Fundamental Theorem of Abelian Groups.
 
  • #9
fresh_42
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Is there anywhere online where a lot of groups are listed along with their unique decomposition according to the Fundamental Theorem of Abelian Groups?
Maybe if I had something like that I could reverse engineer it to figure out how they derived the decompositions.
I have seen some places where various groups are listed along with some groups they are isometric to, but it isn't clear which isometry is the unique decomposition required by the Fundamental Theorem of Abelian Groups.
If we have ##|G|=n## as finite Abelian group and a prime factor decomposition ##n=p_1^{r_1}\cdots p_k^{r_k}## then we get ##G=G_1\times \ldots \times G_k## with ##|G_j|=p_j^{r_j}## in the first step. We can simply sort all elements by their order via ##a\in G_j\backslash \{e\} \Longleftrightarrow p_j|\operatorname{ord}(a).## It should be easy to show that these are subgroups and ##\{e\}## their intersections. Given that, all comes down to distinguish between the possible structures of ##G_j##, i.e. between the groups with ##p^{r}## elements. The question thus is: How can we find the given decomposition ##r=s_1t_1+\ldots+s_mt^m##. And again the orders of the elements gives the answer: Choose an element of maximal order ##p^s## and factor the subgroup generated by that element by the method described above. Then proceed with the factor group.

It can really be done inductively without using complicated propositions. Of course as it is a split into direct factors, we necessarily have to know how ##G=H \times G/H## turns into ##G=H\times K##.
 
  • #10
mathwonk
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this is explained with examples in my notes for math 844, 845

https://www.math.uga.edu/directory/people/roy-smith

In particular in these notes I do (or show how to do) as examples all abelian groups of form (Z/nZ)* = group of units in the ring (Z/nZ).
 
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  • #11
jstrunk
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Thanks. I'll take a look but that seems to be way beyond the level that the book I am using develops this material.
 
  • #12
mathwonk
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You might start by doing the decomposition of cyclic groups, such as Z/(15Z). Do you know the "chinese remainder theorem"?
 
  • #13
mathwonk
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It is a little hard to know what to answer here since the difficulty (and the method) of finding the prime power decomposition of a given abelian (finite) group, depends on how the group is given to you in the first place. The theiorem says of course that a finite abelian group is isomorphic to a product of cyclic groups, each of prime power order. OK. There is also another theorem that says a finite abelian group is isomorphic to a product of cuyclic groups, where the irder of each cyclic factor divides the order of the next one in the product.

So one way of giving the group in the first place is to give a product of the second type, and ask you to express it as a product of the first type. E.g. try expressing Z/(14Z) x Z/(42Z) x Z/(252Z) as a product of cyclic factors each of prime power order. You can make up lots of these examples and solve them. This transition is one basic fact and skill you should have in this subject. Similar examples can be constructed just by starting from any product fo cyclic groups whatever, without regard to how the order of one relates to the order of the next, e.g. Z/(16) x Z/(30Z) x Z/(54Z). Try some of these, reducing them to either a product ofmprime power factors, of a product of factors whose orders do divide each other.

Another way to express a finite abelian group is as a quotient of a free abelian group by a subgroup of maximal rank. All such subgroups are themselves free abelian, so the quotient can always be expressed as the "cokernel" of a homomorphism of free abelian groups, and this homomorphism can always be expressed as a matrix of integers. This is called giving a "presentation matrix" for the quotient group. An m by n matrix of integers defines a homomorphism from Z^n to Z^m, and the image is spanned by the columns. Thus the "cokernel" is the quotient of the target, i.e. of Z^m, modded out by the subgroup spanned by the n columns. In order for this cokernel to be finite, i.e. for the matrix to have maximal rank, there must then be at least m columns, i.e. we must have n ≥ m. Giving a presentation matrix for a group requires finding a finite set of m generators and a finite set of n "relations" among those generators, i.e. linear equations they satisfy. A finite group has a finite set of generators, since just taking all the elements as generators would do, and it is a theorem that any finite set of generators has a finite set of generators for its relations, so every finite abelian group has a presentation matrix.

The easy case for expressing the cokernel as a quotient of cyclic groups is when the matrix is "diagonal", i.e. n ≥ m, the first m columns form a square diagonal m by m matrix, and the rest of the columns are zero. In this case if the jth column has a single non zero entry equal to r in the jth place, this gives rise to a cyclic factor Z/(rZ). So the problem of expressing any group given by any presentation matrix, reduces to the problem of diagonalizing an integer matrix. Thus one proof, and one of the proofs I give in my notes, is to show how to use just the Euclidean algorithm to diagonalize any integer matrix. This expresses your group as a product of cyclic groups, and then you can proceed to change this product into one of the two standard ones mentioned above. Or, one can show that the diagonalizing porocess can be continued in such a way as to give factors whose orders divide each iother consecutively, i.e. one can do the diagonalization so as to give the "second" standard decomposition mentioned above. Then one can change it into a prime power decomposition. Indeed this is one proof I give in my notes.

I did write my notes so that this topic was treated last, after I had covered group theory including the Sylow theorems, but you can perhaps tell from this description here that only the concept of the Euclidean algorithm is needed to carry out the decomposition in practice, at least if you assume as given a presentation matrix. All this is discussed on pages 11-16 of my notes 845-1 on my webpage above. You can see there that the method also applies to finitely generated abelian groups that may not be finite, and gives a decomposition with possibly some factors of Z. This is the case when the matrix may not have maximal rank m. An earlier discussion of cyclicmproducts occurs on pages 47-50 of the notes for 844-2.

And don't be afraid of the words "Sylow p -subgroup". For an abelian finite group, if the group has order np^k where p does not divide n, the sylow p subgroup it is the unique subgroup of order p^k. So in the decomposition, it is just the product of all cyclic factors whose order is a power of p. So all that is happening in these discussions of looking at "Sylow subgroups" is that all factors involving the same prime are being bunched together. E.g. an abelian group G of order 36 = 2^2.3^2, has two Sylow subgroups, one of order 4 and one of order 9, and G is the product of those subgroups, which follows from the "Chinese remainder theorem". I.e. if these subgroups are H and K, the map HxK-->G taking (x,y) to xy, is an isomorphism [exercise].
 
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  • #14
mathwonk
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To emphasize that it matters a lot how the group is given to you in the first place, consider the finite abelian group defined as the multiplicative group of units in the ring Z/(243Z). Try decomposing that into cyclic factors. Hint: It is already cyclic, so your task is just to prove that. This is done in my 844-2 notes, I believe, linked above. By way of contrast, the multiplicative group of units in the ring Z/(32Z) is not cyclic, but has two cyclic factors. The method of proof of these facts is by establishing certain number theoretic results, i.e. it is not a general technique about finite groups. Thus in general, when you ask how to decompose a finite abelian group, the answer again depends on how the group is given to you. In hindsight, this is "obvious", since the task of changing the description of a group from type A to the standard type, of course depends on what type A is. One might claim there is a uniform process for this conversion, starting say from "find an element of maximal order". But of course how to do this depends on how the group is given to you, and trial and error may not be at all practical for a large group - imagine being given a group of order a billion that turns out to be cyclic. The general methods above however give immediately a decomposition of the group of units of the ring Z/(2^60). Or to flog this horse once more, asking how to find the cyclic decomposition of a group is like asking how to get to New York; the answer depends on where you start from. (I started to say asking how to get to Carnegie Hall, but then the answer is the same no matter where you start from, as we all have heard, namely "practice".)
 
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