# Abelian groups of order n.

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.

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at least one million or exactly one million? –  Seth May 8 at 14:39
exactly one million. –  user548793 May 8 at 14:40
mathworld (mathworld.wolfram.com/AbelianGroup.html) gives some formulae for the number of nonisomorphic abelian groups of a given order, however they don't seem to be comfortably usable to solve your question –  Alessandro May 8 at 14:44
Your logic is flawed. For example, there are infinitely many prime numbers but $6$ is not prime. Infinity doesn't mean that everything happens. –  user1729 May 8 at 14:44
Nice variation: show that there is no $n$ such that there are exactly $1000001$ distinct abelian groups of order $n$. –  Dietrich Burde May 8 at 15:07

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.

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Nice-specifically, the smallest possibility is $n=2^4 3^4 5^4 7^4 11^4 13^4 17^2 19^2 23^2 29^2 31^2 37^2=4.9...\times 10^{34}.$ –  Kevin Carlson May 8 at 14:52
I checked that no other partitions work, it has to be 4^6 2^6 1^n for any nonnegative integer n. –  Jack Schmidt May 8 at 14:56

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 $$n = 2^43^45^47^411^413^417^219^223^229^231^237^2 \\= 49,659,789,817,537,838,957,341,175,342,490,000$$

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Yes, the only way, other than superfluous extra primes to the first power, like $n\cdot 43$. –  Jack Schmidt May 8 at 14:57
@JackSchmidt: Which integers are products of partition numbers ? I think, $1000001$ is not. –  Dietrich Burde May 8 at 15:02
@DietrichBurde You can find them yourself here. Sort of. –  Arthur May 8 at 15:03
@DietrichBurde: I wrote a simple backtrack program to find the product decompositions. 10^6+1 is not a product. I believe 13 is the smallest positive integer that is not a product of partition numbers. –  Jack Schmidt May 8 at 15:05
@JackSchmidt Yep, $13$ is the lowest ($1, 2, 3, 5, 7, 11, 15$ goes the partition function, and $13$ is the first prime it misses). –  Arthur May 8 at 15:15

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.

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