Order of abelian groups?

GreenGoblin

Member
"determine all abelian groups of order 1000. which of them contains exactly 3 elements of order two and which of them contain exactly 124 elements of order five?"

ok now I have all the terminology down for this. But its the first time attempting such a type of question. How do I go about this please? I know it has to do with products of primes but I am not sure what exactly its refering to by i.e. 3 elements of order two? what are the elements, is it the primes? how do i find out if i have all the combinations, rather than just repeated trial and error. for example i can make 2^3 * 5^5 = 1000, but how will i know i found all of them? is this even the right procedure? i think so... that it is

Gracias,
GreenGoblin

caffeinemachine

Well-known member
MHB Math Scholar
"determine all abelian groups of order 1000. which of them contains exactly 3 elements of order two and which of them contain exactly 124 elements of order five?"

ok now I have all the terminology down for this. But its the first time attempting such a type of question. How do I go about this please? I know it has to do with products of primes but I am not sure what exactly its refering to by i.e. 3 elements of order two? what are the elements, is it the primes? how do i find out if i have all the combinations, rather than just repeated trial and error. for example i can make 2^3 * 5^5 = 1000, but how will i know i found all of them? is this even the right procedure? i think so... that it is

Gracias,
GreenGoblin
Look up "Fundamental Theorem of finitely Generated Abelian Groups". You will find it in any standard UG text ( eg Herstein's book). From there it wouldn't be difficult.

Deveno

Well-known member
MHB Math Scholar
if G has order 1000, we can write G as H x K, where H has order 8, and K has order 125.

there are 3 possible abelian groups of order 8, namely: Z8, Z4 x Z2, and Z2 x Z2 x Z2, and 3 possible abelian groups of order 125, Z125, Z25 x Z5 and Z5 x Z5 (up to isomorphism, of course). mix and match, that gives us 9 possible abelian groups of order 1000.

it should be clear that elements of order 2 can only live in the "H" part, and the "K" part has to be the identity element of K. so look at how many elements of order 2 are in Z8, Z4 x Z2 and Z2 x Z2 x Z2.

Z8 has but a single element of order 2, namely 4.
every non-identity of Z2 x Z2 x Z2 has order 2, which gives us 7 of them.

that leaves just Z4 x Z2. prove that (2,0), (2,1), and (0,1) are the only elements of order 2, and you're done:

Z4 x Z2 x Z125
Z4 x Z2 x Z25 x Z5
Z4 x Z2 x Z5 x Z5 x Z5 must be the only groups that qualify.

for elements of order 5, use the same reasoning. of our 3 groups of order 125, only Z5 x Z5 x Z5 has 124 elements of order 5 (both Z125 and Z25 x Z5 contain elements of order > 5).

• Amer