HELP Find all abelian groups (up to isomorphism)

[nalkapo]

HELP! Find all abelian groups (up to isomorphism)!

I am really confused on this topic.
can you give me an example and explain how you found, pleaseee!
for example, when i find abelian group of order 20;
|G|=20
i will find all factors and write all of them,
Z_20
(Z_10) * (Z_2)
(Z_5)* (Z_2) * (Z_2)

for higher order such as |G|=200 , i can't do this.(i did and it was wrong)
can you tell me the difference.. how should i find them?
thanks by now...

 

[g_edgar]

Oh, finite ... How boring.
I read the title "Find all abelian groups" so I came here to tell you to read on Ulm's Theorem. But that's not so interesting for finite groups.

 

[nalkapo]

yeah, tell me!
in the textbooks there is nothing.. i can't find any understandable example. just theorems and proofs! is there anyone can help?

 

[VeeEight]

Here is an example I found using google: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2004;task=show_msg;msg=0897.0001
They are making use of the fundamental theorem of finitely generated abelian groups.

 

[nalkapo]

VeeEight, thanks for this example, but it is not explained how they found..

 

[VeeEight]

Here is another link: http://planetmath.org/?op=getobj&from=objects&id=4654

The important thing to remember here is that Z_mn is isomorphic to Z_mxZ_n if and only if gcd (m,n)=1. Therefore, Z₆ is isomorphic to Z₃xZ₂ because gcd(2,3)=1 (and (2)(3)=6). Thus, one way to complete this exercise is to try to decompose your group Z₂₀₀ into a direct product of groups, Z_axZ_bxZ_c... where (a)(b)(c)... = 200 and then apply the above theorem to cancel out the groups that are isomorphic.

For example, 200 can be decomposed as 200=(25)(8). But gcd (25, 8) = 1 so Z₂₅xZ₈ is isomorphic to Z₂₀₀ by the above theorem. Doing this method of writing out all the decompositions and canceling out the isomorphic groups works but can be frustrating since there is a lot of information to work through. Therefore it is helpful to first find the prime factorization of the order (200 in this case) and work out the problem like they did in the link above.

 

[nalkapo]

VeeEight, this answer and the link were really helpful.. Thanks..

 

[nalkapo]

now i get it! by using your information and my homework assignment i got this idea:
for 200 = (2^3).(5^2)
so, our factors can be;
(2^3).(5^2) =Z_8 x Z_25
(2^1).(2^2).(5^2) = Z_2 x Z_4 x Z_25
(2^1).(2^1).(2^1).(5^2)= Z_2 x Z_2 x Z_2 x Z_25
these are for varying exponents of 2.
then,
(2^3).(5^1).(5^1) =Z_8 x Z_25
(2^1).(2^2).(5^1).(5^1) = Z_2 x Z_4 x Z_25
(2^1).(2^1).(2^1).(5^1).(5^1) = Z_2 x Z_2 x Z_2 x Z_25
these are for varying exponents of 5

also, one of the direct product cannot be 200, 100, 50, 40... (Z_200, Z_100, Z_50...), because they are not power of a prime! so, we have only 6 abelian groups up to isomorphism.. Is that true??

 

[Jamma]

All finite abelian groups are of the form (Z_a)x(Z_b)x...x(Z_n) and anything of this form is a finite abelian group.

So if your group is of order 20, it must be one of (Z_20), (Z_10)x(Z_2), (Z_5)x(Z_4), (Z_5)x(Z_2)x(Z_2). But (Z_mn)=(Z_M)x(Z_n) iff m,n are coprime.

So (Z_20) is isomorphic to (Z_5)x(Z_4). (Z_4) is not iso to (Z_2)x(Z_2) because 2 and 2 are not coprime. Also, (Z_20) is not iso to (Z_10)x(Z_2).

So all abelian groups of order 20 are: (Z_20)=iso-to(Z_5)x(Z_4) or (Z_5)x(Z_2)x(Z_2)

Following in this manner, given any number, n, express it as a product of primes, p_1 ... p_m

Then an abelian group of order n is of the form (Z_(p_1))x...(Z_(p_m)) except that if we have repeated primes, we may multiply them together, to give a different group which has some factors (Z_(p_i)^j) say. (We can do this for any any j and any of the p_i's)

So with 20, 20=2x2x5 is a prime factorisation of 20. So we have either (2,2,5) or (4,5) as we had before.

Hope this makes things simpler.

 

[seshikanth]

I too have a confusion regarding this problem:
Question 1:
------------
"So all abelian groups of order 20 are: (Z_20)=iso-to(Z_5)x(Z_4) or (Z_5)x(Z_2)x(Z_2)"
Here the order of (Z_5)x(Z_2)x(Z_2) is lcm(5,2,2) = 10. As the isomorphism is order preserving, Z_20 being order 20 and (Z_5)x(Z_2)x(Z_2) being order 10 are not isomorphic,. Please correct me if i am wrong here or missing something!

Question 2:
------------
Also, in "First Course in Abstract Algebra by Fraleigh" Example:11.13: Find all abelian groups upto isomorphism of order 360?
Solution for this example is given as:
1. (Z_2)x(Z_2)x(Z_2)x(Z_3)x(Z_3)x(Z_5) ----> 2^3 * 3^2 *5 ---- order = lcm(2,3,5) = 30
and then he proceeds with prime Factorization of n = 360.
As this (Z_2)x(Z_2)x(Z_2)x(Z_3)x(Z_3)x(Z_5) being order 30, how can it be listed under Finite abelian groups of order 360.

Thanks,

 

[quasar987]

I too have a confusion regarding this problem:
Question 1:
------------
"So all abelian groups of order 20 are: (Z_20)=iso-to(Z_5)x(Z_4) or (Z_5)x(Z_2)x(Z_2)"
Here the order of (Z_5)x(Z_2)x(Z_2) is lcm(5,2,2) = 10. As the isomorphism is order preserving, Z_20 being order 20 and (Z_5)x(Z_2)x(Z_2) being order 10 are not isomorphic,. Please correct me if i am wrong here or missing something!

You are wrong or missing something. Look at Z_2 x Z_3 for example. What are its elements?
First fix the first factor to 0. The elements of this kind are (0,0), (0,1), (0,2). Now fix the second factor to 1. The elements of this kind are (1,0), (1,1), (1,2). For a total of 3+3 = 2 x 3 = 6. You can generalize this argument to see that Z_n x Z_m x ... x Z_l has n x m x ... x l elements, and not lcm(n, m, ..., l).

 

[seshikanth]

Hi,
In the example above Z2 x Z3 has order 6. Note that order is minimum smallest integer sothat any element in the Group goes to identity element. In Number of elements, m*n*o*p*... , there may be elements repetitive. For example., Group (Cartesian product) Z4 * Z2 has order 4. Because (1,1) + (1,1) ... (4 times ) .. = (0,0)
But the product 4*2 = 8elements has repetitions due to modular arithmetic.

So, the order of Zm * Zn * Zo * ... = lcm(m,n,o,p...)

THanks,

 

[seshikanth]

hmmm... ok i got it!

 

[quasar987]

The order of a group is just it's cardinality.

Not to be confused with the order of an element of the group.

 

[Jamma]

Yep, you were just confusing your definition of the order of a group. Given that we already have a word for the number of elements in a group, it's cardinality, maybe your definition makes more sense

If we were using this definition, then for an integer n I guess that you would just have all groups of the form Z_n x Z_(x_1) x Z_(x_2) x ... x Z_(x_k) where the x_i are all divisors of n.