Question

I've been asked to find all the isormphic classes of abelian groups of order 760. I google searched and found for 720:
Z2×Z2×Z2×Z2×Z3×Z3×Z5
Z4×Z2×Z2×Z3×Z3×Z5
Z4×Z4×Z3×Z3×Z5
Z8×Z2×Z3×Z3×Z5
Z16×Z3×Z3×Z5
Z16×Z9×Z5
Z8×Z2×Z9×Z5
Z4×Z4×Z9×Z5
Z4×Z2×Z2×Z9×Z5
Z2×Z×Z2×Z2×Z9×Z5

How would 760 look?

Answer 1

Abelian groups are relatively to describe. The only thing you need to know is the prime factorization of the order. In this case, we have that \(760=2^3\cdot5\cdot19\). Since finite abelian groups can be written as a direct product of cyclic groups, all you need to to do is realize how many ways you can split \(2^3\cdot5\cdot19\) into factors that are NOT coprime.

Answer 2

ah ok why are there fewer than 720

Answer 3

There are much fewer because the prime powers are smaller. 720 factors as\[720=2^4\cdot3^2\cdot5\]In this factorization you have both a power of 4 and 2, but in the factorization of 760, we only had a power of 3.

Answer 4

A general formula for the number of abelian groups of order\[n=p_1^{e_1}\cdot p_2^{e_2}\cdot...\cdot p_r^{e_r}\]is given by\[f(e_1)\cdot f(e_2)\cdot...\cdot f(e_r)\]where \(f(e_i)\) is defined to be the number of integer partitions of \(e_i\). See this article for a description of integer partitions. http://en.wikipedia.org/wiki/Partition_(number_theory)

Answer 5

Ok thank you Im have a look

Answer 6

* im gonna have a look

Answer 7

Oh, and just to be clear, that factorization of \(n\) I gave is the unique prime factorization.

Answer 8

Sorry, I think I confused myself a little bit when writing the groups of order 760. There are actually 3 groups of that order, and they're actually\[\mathbb{Z}_{760}\]\[\mathbb{Z}_2\times\mathbb{Z}_{380}\]\[\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_{190}\]When I said you need to look at the factors that are not coprime, I left out the additional restriction that each factor need to be divisible by the previous factor.

Answer 9

So in particular, if you have an abelian group of order \(p_1^{e_1}\cdot...\cdot p_{r}^{e_r}\), then you will always have at least one subgroup of order \(p_1\cdot...\cdot p_r\).

Answer 10

interesting.. still a little confused with this stuff but you cleared it up for me thanks!

Answer 11

You're welcome. Feel free to ask me any more questions on this stuff.

Answer 12

Will do!