The classification of finite abelian groups

 Stated without proof: Every finite abelian group is the direct product of a finite number of cyclic groups of the form Zn.

 

If I make up an arbitrary finite list of cyclic groups, it's a finite abelian group. For example,

 

 Z5 ✕ Z14 ✕ Z2 ✕ Z17 ✕ Z44 ✕ Z10 

 

The order of this group is 5 ✕ 14 ✕ 2 ✕ 17 ✕ 44 ✕ 10 = 1,057,200. Because this number is not prime, there are a lot of groups of order 1,057,200, and we need a way to tell them apart and a consistent way to represent them. Mathematicians have decided on two consistent ways to write down a group like this, primary decomposition and cyclic decomposition.

 

First step: write each component subgroup as a direct product of groups whose order are powers of primes.

 

We have three components that are already powers of primes, so I will put them at the front of the list in numerical order.

 Z2 ✕ Z5 ✕ Z17 ✕ Z14 ✕ Z44 ✕ Z10

Now we split of the composite numbers into powers of primes.

Z14 = Z2 ✕ Z7  

Z44 = Z4 ✕ Z11 

Z10 = Z2 ✕ Z5

Notice that Z44 is not Z2 ✕ Z2 ✕ Z11. Z4 has a single generating element, Z2 ✕ Z2 but needs two generators. Now we replace the last three groups with their primary decompositions.

 

Z2 ✕ Z5 ✕ Z17 ✕ Z2 ✕ Z7 ✕ Z4 ✕ Z11 ✕ Z2 ✕ Z5 

 

We put them in numerical order with this proviso: powers of 2 first with highest power at the front of that list, then powers of 3, powers of 5, powers of 7, etc. There are no powers of 3 in the list, so we skip that prime. I will use parentheses to separate the lists of different primes from one another.

(Z4 ✕ Z2 ✕ Z2 ✕ Z2) ✕ (Z5 ✕ Z5) ✕ (Z7) ✕ (Z11) ✕ (Z17)

 

This is the primary decomposition of our arbitrary group.

 

For the cyclic group, we take the representatives of the primes in order and multiply them together. I have marked the numbers in red.

 

(Z4 ✕ Z2 ✕ Z2 ✕ Z2) ✕ (Z5 ✕ Z5) ✕ (Z7) ✕ (Z11) ✕ (Z17)

 

4 ✕ 5 ✕ 7 ✕ 11 ✕ 17 = 26,180

 

Next: We multiply the second group orders in each primes' list, here marked in blue.

 

(Z4 ✕ Z2 ✕ Z2 ✕ Z2) ✕ (Z5 ✕ Z5) ✕ (Z7) ✕ (Z11) ✕ (Z17)

 

2 ✕ 5 = 10

 

Now we just have the two remaining copies of Z2. The list {26180, 10, 2, 2} has the property that each number is a multiple of the number that comes after it. So the cyclic decomposition is

 

Z26180 ✕ Z10 ✕ Z2 ✕ Z2 

 

One nice thing about cyclic decomposition is we know largest subgroup generated by a single element has order 26,180. Another point in its favor is it the direct product of just four groups instead of the nine groups we needed for the primary decomposition.