Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?

Question

Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?

Can anyone please explain. I would yes because numbers are infinitive, and so any number $n$ can be expressed as a direct product of cyclic groups of order $n$.

Answer 1

By the fundamental theorem for finite abelian groups the number of abelian groups of order $n=p_1^{n_1}\dots p_k^{n_k}$ is the product of the partition numbers of $n_i$.

Note that the partition number of $2$ is $2$ and the partition number of $4$ is $5$. Since $10^6=2^6\cdot 5^6$ such an $n$ therefore exists.

Answer 2

Given any number $n$, the number of different (up to isomorphism) abelian groups of order $n$ is dependant on the prime factorization of $n$. If $n$ is a prime power $p^k, k\in \Bbb N$, then the number of abelian groups of order $n$ is exactly the number of distinct partitions of $k$. Example: For $p = 3, k = 4$ and thus $n = 81$, there are $5$ different abelian groups: $$ \Bbb Z_{81} \quad \Bbb Z_{27}\times \Bbb Z_3 \quad \Bbb Z_{9} \times \Bbb Z_9 \quad \Bbb Z_9 \times \Bbb Z_3 \times \Bbb Z_3 \quad (\Bbb Z_3)^4 $$ Now, if there are multiple primes in the factorization of $n$, the number of partitions allowed by each one are just multiplied together. Example: If $n = 2^53^7 = 69,984$, then there are $7\cdot 15 = 105$ different abelian groups, since $5$ has $7$ partitions and $7$ has $15$.

Now, to construct a million, we need $2^65^6$ different partitions total. The easiest way (maybe the only way) is to let $n$ have be $6$ primes to the power $2$ ($2$ partitions) and $6$ primes to the power $4$ (five partitions). Thus, I believe the smallest such $n$ is given by $$ n = 2^43^45^47^411^413^417^219^223^229^231^237^2 \\= 49,659,789,817,537,838,957,341,175,342,490,000 $$

Answer 3

If $n=\prod_{i=1}^rp_i^{k_i}$, then the number of distinct abelian groups of order $n$ is given by $$ \prod_{i=1}^rp(k_i), $$ where $p(k)$ denotes the number of partitions of $k$. Now $10^6=p(k_1)\cdots p(k_r)$ has to be solved. But $p(2)=2$ and $p(4)=5$, so we can solve it.

Answer 4

In order to get all of the abelian groups of order $n = \prod\limits_{i=1}^s p_i^{a_i}$ you must first get each possibility for each prime power, and then take all of the combinations.

There are $f(a)$ abelian groups of order $p^a$ where $f$ is the partition function and they are given by the distinct partitions of the exponent. The number of abelian groups of order $n = \prod\limits_{i=1}^s p_i^{a_i}$ is $\prod\limits_{i=1}^s f(a_i)$.

In our case we have $n=2^2\cdot5^2\cdot 7^2$ so there are $f(2)^3 = 2^3 = 8$ abelian groups.

The possibilities are:

$\mathbb Z_4 \times \mathbb Z_{25} \times \mathbb Z_{49}$

$\mathbb Z_4 \times \mathbb Z_{25} \times \mathbb Z_{7}\times \mathbb Z_{7}$

$\mathbb Z_4 \times \mathbb Z_{5} \times \mathbb Z_{5} \times \mathbb Z_{49}$

$\mathbb Z_4 \times \mathbb Z_{5} \times \mathbb Z_{5} \times \mathbb Z_{7} \times \mathbb Z_7$

$\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_{25} \times \mathbb Z_{49}$

$\mathbb Z_2 \times \mathbb Z_2\times \mathbb Z_{25} \times \mathbb Z_{7}\times \mathbb Z_{7}$

$\mathbb Z_2 \times \mathbb Z_2\times \mathbb Z_{5} \times \mathbb Z_{5} \times \mathbb Z_{49}$

$\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_{5} \times \mathbb Z_{5} \times \mathbb Z_{7} \times \mathbb Z_{7}$

As you can see the the first half have $\mathbb Z_4$, and in each of the halves, the first half of the half has $\mathbb Z_{25}$, and finally, the ones in odd position are the ones that have $\mathbb Z_{49}$, so we are just taking all the combinations. Also, this result is called the fundamental theorem of finite abelian groups.