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$67500=2^2*3^3*5^4$ => $2*3*4=24$ pairwise non-isomorphic abelian groups.

Is this correct?

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2 Answers 2

up vote 3 down vote accepted

No you don't multiply the exponents but the product of the numbers of integer partitions of the exponents.

As $2=1+1$ and $2=2$ and $3=1+2=1+1+1$ and $4=1+1+1+1=2+1+1=2+2=3+1$ you have $$ 2\cdot 3 \cdot 5=30$$ non isomorphic abelian groups.

This is because of the classification theorem for finite abelian groups states, that when $G$ is finite and abelian, then $$G\cong \bigoplus_{i=1}^n \mathbb{Z}/p_i^{v_i}\mathbb{Z}$$

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Are you sure its correct? For example $18=2^1*3^2$, by your method it would have $6$ non isomorphic abelian groups. But it has only $2$ of them. –  blondy Sep 5 '13 at 11:08
    
@blondy You should try to read it again. The actual primes involved are unimportant. –  Tobias Kildetoft Sep 5 '13 at 11:12
    
@Blondy, what Tobias says is that If $n = \prod p_i^{k_1}$ and $\pi(k)$ denotes the number of integer partitions of $k$, then the number of distinct abelian groups of order $n$ is $\prod \pi(k_i)$. –  Nicky Hekster Sep 5 '13 at 11:18
    
Sorry I am still not getting it. Where is the $2*3*5$ coming from? Precisely where is the $5$ coming from? –  blondy Sep 5 '13 at 11:18
    
Oh wait, now I am getting it. $4$ can be partitioned in five ways. $4, 1+1+1+1,2+2,2+1+1,3+1$. –  blondy Sep 5 '13 at 11:22

By the Classification Theorem for Finite Abelian Groups, all finite abelian groups are isomorphic to $$\mathbb{Z}_{q_1} \times \mathbb{Z}_{q_2} \times \cdots \times \mathbb{Z}_{q_t}$$ where $q_1,\ldots,q_t$ are $t$ not necessarily distinct prime powers.

Consequently, the number of finite abelian groups of order $p^k$ for some prime $p$ is the number of partitions of $k$. For example, the number of groups of order $p^4$, for prime $p$, are tabulated below.

$$ \begin{array}{rr} \text{partition} & \text{abelian group} \\ 4 & \mathbb{Z}_{p^4} \\ 3 + 1 & \mathbb{Z}_{p^3} \times \mathbb{Z}_{p} \\ 2 + 2 & \mathbb{Z}_{p^2} \times \mathbb{Z}_{p^2} \\ 2 + 1 + 1 & \mathbb{Z}_{p^2} \times \mathbb{Z}_{p} \times \mathbb{Z}_{p} \\ 1 + 1 + 1 + 1 & \mathbb{Z}_{p} \times \mathbb{Z}_{p} \times \mathbb{Z}_{p} \times \mathbb{Z}_{p} \\ \end{array}$$

We can also deduce that if $p$ and $q$ are distinct primes, then the abelian groups of order $p^k q^r$ are isomorphic to $G \times H$ where $G$ is an abelian group of order $p^k$ and $H$ is an abelian group of order $q^r$.

So, if $\pi_n$ denotes the number of partitions of $n$, we have:

  • The number of abelian groups of order $p^k$ for prime $p$ is $\pi_k$.

  • The number of abelian groups of order $p^k q^r$ for distinct primes $p$ and $q$ is $\pi_k \pi_r$.

  • And so on.

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